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Questions and Answers
What does the negative value of the total energy of an electron in an orbit signify, according to Bohr's theory?
What does the negative value of the total energy of an electron in an orbit signify, according to Bohr's theory?
- The electron is not bound to the nucleus and can escape freely.
- The electron is bound to the nucleus by electrostatic attraction, requiring work to be done to remove it. (correct)
- The electron possesses excess kinetic energy.
- The electron's potential energy is zero.
Based on the relationship $v = \frac{c}{137n}$, how does the velocity of an electron in the hydrogen atom change with increasing orbit number $n$?
Based on the relationship $v = \frac{c}{137n}$, how does the velocity of an electron in the hydrogen atom change with increasing orbit number $n$?
- The velocity increases exponentially with $n$.
- The velocity increases proportionally with $n$.
- The velocity remains constant regardless of $n$.
- The velocity decreases proportionally with $n$. (correct)
For the Lyman series, if an electron transitions from $n_2 = 3$ to $n_1 = 1$, how would you calculate the wave number $\bar{\nu}$ using Rydberg's constant R?
For the Lyman series, if an electron transitions from $n_2 = 3$ to $n_1 = 1$, how would you calculate the wave number $\bar{\nu}$ using Rydberg's constant R?
- $\bar{\nu} = R \left[ \frac{1}{1^2} - \frac{1}{3^2} \right]$ (correct)
- $\bar{\nu} = R \left[ \frac{1}{1} - \frac{1}{3} \right]$
- $\bar{\nu} = R \left[ \frac{1}{3^2} - \frac{1}{1^2} \right]$
- $\bar{\nu} = R \left[ \frac{1}{3} - \frac{1}{1} \right]$
What region of the electromagnetic spectrum does the Lyman series of the hydrogen atom fall into?
What region of the electromagnetic spectrum does the Lyman series of the hydrogen atom fall into?
Using the given formula $\alpha = \frac{2 \pi \times 9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{3 \times 10^{8} \times 6.63 \times 10^{-34}}$, what physical quantity does $\alpha$ represent and what is its approximate value?
Using the given formula $\alpha = \frac{2 \pi \times 9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{3 \times 10^{8} \times 6.63 \times 10^{-34}}$, what physical quantity does $\alpha$ represent and what is its approximate value?
What is the fundamental principle behind the photoelectric effect?
What is the fundamental principle behind the photoelectric effect?
Which of the following statements accurately describes the relationship between the intensity of incident light and the photoelectric current?
Which of the following statements accurately describes the relationship between the intensity of incident light and the photoelectric current?
What is the significance of the stopping potential in the context of the photoelectric effect?
What is the significance of the stopping potential in the context of the photoelectric effect?
How does the stopping potential vary with the frequency of incident radiation?
How does the stopping potential vary with the frequency of incident radiation?
What does the formula $K_{max} = rac{1}{2} m v_{max}^{2} = e V_{0}$ represent in the photoelectric effect?
What does the formula $K_{max} = rac{1}{2} m v_{max}^{2} = e V_{0}$ represent in the photoelectric effect?
Why are alkali metals like Lithium, Sodium, and Potassium highly photosensitive?
Why are alkali metals like Lithium, Sodium, and Potassium highly photosensitive?
For two different metals, A and B, the graphs of stopping potential versus frequency of incident radiation are parallel straight lines. What does this indicate?
For two different metals, A and B, the graphs of stopping potential versus frequency of incident radiation are parallel straight lines. What does this indicate?
Which type of electromagnetic radiation is most likely to eject electrons from heavy metals?
Which type of electromagnetic radiation is most likely to eject electrons from heavy metals?
A metal with a work function of 3.0 eV is illuminated with light of frequency $8.0 imes 10^{14}$ Hz. What is the maximum kinetic energy of the emitted photoelectrons?
A metal with a work function of 3.0 eV is illuminated with light of frequency $8.0 imes 10^{14}$ Hz. What is the maximum kinetic energy of the emitted photoelectrons?
If the stopping potential for a certain metal is 1.5 V when illuminated by light of wavelength 500 nm, what is the work function of the metal?
If the stopping potential for a certain metal is 1.5 V when illuminated by light of wavelength 500 nm, what is the work function of the metal?
Light of a certain frequency is incident on a metal surface, and photoelectrons are emitted. If the intensity of the light is doubled, what happens to the stopping potential?
Light of a certain frequency is incident on a metal surface, and photoelectrons are emitted. If the intensity of the light is doubled, what happens to the stopping potential?
Consider two metals, A and B, with different work functions, $W_A$ and $W_B$, where $W_A > W_B$. If both metals are illuminated with light of the same frequency, which is greater than both threshold frequencies, which metal will have a higher stopping potential?
Consider two metals, A and B, with different work functions, $W_A$ and $W_B$, where $W_A > W_B$. If both metals are illuminated with light of the same frequency, which is greater than both threshold frequencies, which metal will have a higher stopping potential?
An electron has a de Broglie wavelength of 0.1 nm. What is its momentum?
An electron has a de Broglie wavelength of 0.1 nm. What is its momentum?
A photon and an electron have the same de Broglie wavelength. Which of the following is true regarding their momentum?
A photon and an electron have the same de Broglie wavelength. Which of the following is true regarding their momentum?
What is the de Broglie wavelength of an electron with a kinetic energy of 100 eV?
What is the de Broglie wavelength of an electron with a kinetic energy of 100 eV?
If the kinetic energy of an electron is doubled, by what factor does its de Broglie wavelength change?
If the kinetic energy of an electron is doubled, by what factor does its de Broglie wavelength change?
Flashcards
Photoelectric Effect
Photoelectric Effect
Emission of electrons from a metal surface when electromagnetic radiation of sufficient frequency hits it.
Photoelectrons
Photoelectrons
Electrons emitted from a metal surface due to the photoelectric effect.
Intensity of Light
Intensity of Light
Current is directly proportional to the number of photoelectrons emitted per second.
Stopping Potential
Stopping Potential
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K_max Equation
K_max Equation
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Stopping Potential vs. Frequency
Stopping Potential vs. Frequency
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Threshold Frequency
Threshold Frequency
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Stopping Potential graphs for Metals
Stopping Potential graphs for Metals
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Photon Energy
Photon Energy
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Photons per second
Photons per second
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Wavelength Calculation
Wavelength Calculation
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De Broglie Waves
De Broglie Waves
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Einstein's Mass-Energy Relation
Einstein's Mass-Energy Relation
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De Broglie Wavelength
De Broglie Wavelength
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Matter Waves
Matter Waves
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Fine-structure constant ($\alpha$)
Fine-structure constant ($\alpha$)
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Electron velocity (v) in first orbit
Electron velocity (v) in first orbit
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Kinetic Energy (K.E) in nth orbit
Kinetic Energy (K.E) in nth orbit
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Potential Energy (P.E) in nth orbit
Potential Energy (P.E) in nth orbit
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Lyman Series
Lyman Series
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Study Notes
- Final marathon for Class 12th physics covering modern physics with Shailendra Pandey Sir.
- Join Shailendra Sir's official Telegram channel for PDF notes accessible via the QR code.
Photoelectric effect
- The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of high frequency hits it.
- The emitted electrons are called photoelectrons.
- Alkali metals like Lithium (Li), Sodium (Na), Potassium (K), Cesium (Cs), and Rubidium (Rb) are very photosensitive and emit electrons even with visible light.
- Metals like Zinc (Zn), Cadmium (Cd), Magnesium (Mg), and Aluminum (Al) only respond to ultraviolet light.
- X-rays can eject electrons from even heavy metals.
Experimental study of the photoelectric effect: Intensity
- The photoelectric current is directly proportional to the number of photoelectrons emitted per second.
- The number of photoelectrons emitted is proportional to the intensity of the incident radiation.
Experimental study of the photoelectric effect: Potential
- The retarding potential at which the photoelectric current becomes zero is the cut-off or stopping potential.
Experimental study of the photoelectric effect: Frequency
- The stopping potential increases with the frequency of the incident radiation.
- For frequencies V3 > V2 > V1, the corresponding stopping potentials vary as V03 > V02 > V01.
- The stopping potential increases linearly with the frequency of the incident radiation for a given photosensitive material.
- The graphs for two different metals A and B are parallel straight lines with the same slope.
- Threshold frequencies differ for the metals.
- hv = W + Kmax (where Kmax is the kinetic energy)
- hv = Wo + eV0 (where Wo is the is work function and eV0 is the energy)
- eV0 = hv - Wo
Question: Monochromatic light from a laser
- A laser produces monochromatic light with a frequency of 6.0 x 10^14 Hz and emits power of 2.0 x 10^-3 W.
- The energy of each photon is 3.98 x 10^-19 J.
- The source emits 5.0 x 10^15 photons per second.
- P=N x energy of each photon.
Work function question
- The work function of cesium is 2.14 eV.
- The threshold frequency for cesium if the photocurrent is brought to zero by stopping potential to 0.60 eV is 5.16 x 10^14 Hz.
- The wavelength of the incident light is approximately 453.7 nm
Dual nature of matter: De Broglie waves
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Matter or de Broglie waves are waves associated with the material particles, and their wavelength is called the de Broglie wavelength.
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Photon energy can be described with Einstein's mass-energy relationship as E = mc².
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λ = h/p = h/mc
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K.E. gained by the electron can be expressed as K = (1/2)mv² = p²/2m, leading to p =√(2mK).
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De Broglie wavelength is represented by λ = h/p = h/√2mK = h/√2meV.
De Broglie wavelength of an electron
- λ = h/p = h/√(2mK) = h/√(2meV)
- For an electron given h = 6.63 x 10⁻³⁴ Js, m = 9.1 x 10⁻³¹ kg, and e = 1.6 x 10⁻¹⁹ C :
- λ = (12.3/√V) Å
Failure of Thomson's model
- Thomson's model was discarded because it failed to explain the origin of spectral series in hydrogen.
- It could not explain the large angle scattering of alpha particles in Rutherford’s experiment.
Rutherford's alpha scattering experiment
- Most alpha particles passed through the gold foil without significant deflection.
- Different alpha particles underwent varying degrees of deflection.
- Few alpha particles (about 1 in 8,000) retraced their paths or deflected nearly 180°.
Distance of closest approach
- The distance of the closest approach is denoted by r₀ = (1 / 4πε₀) (2Ze²/Kα).
- Where Kα is the kinetic energy of alpha particle.
Impact parameter
- The impact parameter is the perpendicular distance of the velocity vector of an alpha particle to the center of the nucleus when far from the atom.
- b = (1 / 4πε₀) (Ze²cot(θ/2) / (1/2)mu²)
Drawbacks of Rutherford's atom model
- Contradicted electromagnetic theory, predicting continuous energy loss and atom collapse.
- The Rutherford model could not explain the stability of an atom.
- Failed to explain the discrete line spectrum of hydrogen, because it said the electron should revolve in orbits of all possible radii.
Calculate impact parameter
- Given K = 5MeV = 5 x 1.6 x 10⁻¹³J and θ = 90°, Z = 79, the impact parameter is b = 2.27 x 10⁻¹⁴m.
Bohr's quantisation condition
- According to de Broglie, electrons are associated with wave characteristics, so a circular orbit will be a stationary energy only it contains complete de-Broglie wavelengths
- 2πr = nλ (where n is an integer)
- Angular momentum (L) of the electron must be L=mvr=nh/2π, with n = 1, 2, 3...
- Only circular orbits are allowed where an electrons angular momentum is a multiple of h/2π.
Bohr's theory of hydrogen atom
- Kinetic Energy (KE) = Kze²/2r
- Potential Energy (PE) = -Kze²/r
- Total Energy (TE) =-Kze²/2r
- r = n²h² / 4π²mKze²
- v = 2πkze² / nh
- Bohr radius: r = (n²/Z) * (0.526 Å)
- Total energy levels of hydrogen-like atoms: TE = -13.6 eV (Z²/n²)
Bohr's theory for hydrogen
- Equation: v=2πKZe²/nh = (2πKe²/ch) *(c/n) , which is also v=α * (c/n) where alpha = (2πKe²/ch). Therefore, a = 1/137
- For the first orbit(n=1), v = c/137
Bohr's theory Kinetic/Potential Energy
- Kinetic energy = 1/2mv² = kZe²/2r
- Potential energy = k(q1q2)/r = kZe(-e)/r = kZe²/r
- Total Energy = En = -(2π²mk²Z²e⁴) / (n²h²)
- Negative total energy indicates the electron is bound to nucleus by electrostatic attraction
Energy level transitions question
- Energy Level Example: Given ground state energy E₁ = -13.6 eV:
- E₄ energy transition: λ photon energy (hν) will be produced.
Spectral series of hydrogen atom
- Describes the different series based on electron transitions between energy levels, 1/λ = R[1/n₁² - 1/n₂²]
- Lyman series (UV Region)
- Electrons jump to n₁ = 1 from n₂ = 2, 3, 4,...
- Balmer series (Visible region)
- transition from n₂=3/4/5 to n₁=2
- Paschen series
- if n₂ = 4,5,6...and n₁ = 3, we get a spectral series in infrared region, paschen series.
- Brackett series
- when n₂ = 5, 6, 7... and n₁ = 4 results in Brackett series( infrared region)
- Pfund series
- if n₂=6,7,8... and n₁=5, a spectral series in the infrared emerges, known as Pfund series.
Energy diagram hydrogen
- Depicts different energy levels hydrogen atom represented in eV.
- Features UV region where Lyman series occurs when it transitions -13eV to -3.4(n=2), -1.51(n=3), -.85ev (n=4)
- There is a visible region is seen in 3.4eV, known as Balmer Region. Lastly, Infrared can be from -.85ev and lower.
Limitations Bohr's Theory
- Applicable to Hydrogen- similar single electron atoms, cannot work with multiple electron atoms
- Cannot explain fine features Hydrogen spectrum where groups closed lines w/ slightly different frequencie
- Only explains circular orbits, not elliptical
- Electons exhibit wave properties as well, therefore cannot be defined as in Bohr's Theory
- Does not provide information relative intensities, only frequencies or lines Does not describe spectral split that occurs electric and magnetic Fields
Excitation energy
- Excitation energy: energy required for electron to jump from ground state to higher, excited states. First excitation: E2 to E1=-3.4eV-(-13.6) eV =10.2eV
Ionization Energy
- Ionisation Energy It refers amount energy require electron knock from atom completely, ground or outermost(n =∞
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Description
Explore quantum numbers, atomic structure, and the photoelectric effect. Understand the significance of negative total energy, electron velocity in hydrogen atoms, and Lyman series calculations. Investigate the relationship between light intensity and photoelectric current.