When the light of frequency 2u0 (where u0 is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident... When the light of frequency 2u0 (where u0 is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5u0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is: Read more

Steps to Solve

1. Understanding the photoelectric effect The kinetic energy of the emitted electrons can be determined using the photoelectric effect equation: $$ KE = E - \phi $$ where ( KE ) is the kinetic energy, ( E ) is the energy of the incident photon given by ( E = h f ), and ( \phi ) is the work function (related to the threshold frequency ( u_0 )).

2. Calculate velocity from kinetic energy The kinetic energy can also be expressed in terms of the maximum velocity ( v ) of the electrons: $$ KE = \frac{1}{2} m v^2 $$ where ( m ) is the mass of the electron.

3. Expressing the energies for different frequencies For the first frequency ( 2u_0 ):

• The energy of the photon is: $$ E_1 = h \cdot 2u_0 $$
• The kinetic energy of emitted electrons is: $$ KE_1 = E_1 - \phi = h \cdot 2u_0 - h u_0 = h u_0 $$

For the second frequency ( 5u_0 ):

• The energy of the photon is: $$ E_2 = h \cdot 5u_0 $$
• The kinetic energy of emitted electrons is: $$ KE_2 = E_2 - \phi = h \cdot 5u_0 - h u_0 = 4h u_0 $$

4. Finding the expressions for velocities For ( v_1 ) from ( KE_1 ): $$ \frac{1}{2} m v_1^2 = h u_0 $$ Thus, $$ v_1^2 = \frac{2h u_0}{m} $$ Taking the square root gives: $$ v_1 = \sqrt{\frac{2h u_0}{m}} $$

For ( v_2 ) from ( KE_2 ): $$ \frac{1}{2} m v_2^2 = 4h u_0 $$ Thus, $$ v_2^2 = \frac{8h u_0}{m} $$ Taking the square root gives: $$ v_2 = \sqrt{\frac{8h u_0}{m}} $$

5. Finding the ratio of velocities Now, we can find the ratio ( \frac{v_1}{v_2} ): $$ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2h u_0}{m}}}{\sqrt{\frac{8h u_0}{m}}} $$

This simplifies to: $$ \frac{v_1}{v_2} = \frac{\sqrt{2}}{\sqrt{8}} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} $$