1.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? 1.4 Monochromatic light of wavelength 632.8 nm... 1.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? 1.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Understand the Problem

The question asks for calculations related to the photoelectric effect and properties of photons emitted from a helium-neon laser. It involves finding the maximum kinetic energy of photoelectrons, energy and momentum of a photon, the number of photons emitted per second, and the speed required for a hydrogen atom to have a specific momentum.

Answer

Maximum kinetic energy: $ KE_{max} = 2.4 \times 10^{-19} \, J $, Energy per photon: $ E = 3.14 \times 10^{-19} \, J $, Number of photons per second: $ N \approx 30.0 $, Speed of hydrogen atom: $ v = 6.38 \times 10^{6} \, m/s $

Steps to Solve

Calculate the maximum kinetic energy of photoelectrons

The maximum kinetic energy ($KE_{max}$) of photoelectrons is given by the formula: $$ KE_{max} = eV $$ where ( e ) (the charge of an electron) is approximately ( 1.6 \times 10^{-19} ) C, and ( V ) is the cutoff voltage in volts.

Substituting the values: $$ KE_{max} = (1.6 \times 10^{-19} , C)(1.5 , V) = 2.4 \times 10^{-19} , J $$

Calculate the energy of each photon

The energy of a photon ($E$) can be calculated using the formula: $$ E = \frac{hc}{\lambda} $$ where ( h ) is Planck's constant (( 6.626 \times 10^{-34} , J \cdot s )), ( c ) is the speed of light (( 3.00 \times 10^8 , m/s )), and ( \lambda ) is the wavelength in meters.

Convert ( \lambda ) from nm to m: $$ \lambda = 632.8 , nm = 632.8 \times 10^{-9} , m $$

Now calculate ( E ): $$ E = \frac{(6.626 \times 10^{-34} , J \cdot s)(3.00 \times 10^8 , m/s)}{632.8 \times 10^{-9} , m} $$

Calculate the momentum of each photon

Use the formula for the momentum ($p$) of a photon: $$ p = \frac{E}{c} $$ The value of ( E ) is from the previous step. Substitute to find ( p ).

Calculate the number of photons emitted per second

The number of photons emitted per second ($N$) can be found using the formula: $$ N = \frac{P}{E} $$ where ( P ) is the power of the laser (in watts) and ( E ) is the energy of a single photon (from step 2).

Calculate the speed of a hydrogen atom with the same momentum as the photon

The momentum of the hydrogen atom ($p_{H}$) must equal the momentum of the photon: $$ p_{H} = mv $$ where ( m ) is the mass of the hydrogen atom (( 1.67 \times 10^{-27} , kg )) and ( v ) is its speed.

Rearranging this gives: $$ v = \frac{p}{m} $$ Substituting the value of ( p ) from step 3 and the mass of the hydrogen atom.

Maximum kinetic energy of photoelectrons: ( KE_{max} = 2.4 \times 10^{-19} , J )
Energy of each photon: $$ E = 3.14 \times 10^{-19} , J $$
Momentum of each photon: $$ p = 1.05 \times 10^{-36} , kg \cdot m/s $$
Number of photons emitted per second: $$ N \approx 30.0 , \text{photons/second} $$
Speed of a hydrogen atom with equal momentum: $$ v = 6.38 \times 10^{6} , m/s $$

More Information

The photoelectric effect illustrates the particle nature of light, demonstrating that light can be thought of as particles (photons) that can impart energy to electrons. The helium-neon laser emits photons that interact with electrons, resulting in the photoelectric effect. The calculations also show how physical constants are involved in deriving energies and momenta.

Tips

Incorrect unit conversions: Always ensure wavelengths are in meters when calculating photon energy.
Confusing power and energy: Remember that power is energy per unit time and must be carefully distinguished in calculations.
Forgetting to use consistent units: Ensure all values are in the same system (e.g., SI) before calculations.

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