Question

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

• A. 2:01
• B. 1:02
• C. 4:01
• D. 1:04

Answer 1

The Correct Option is B

To solve this problem, we need to use the concept of the photoelectric effect. According to Einstein's photoelectric equation, the kinetic energy of emitted electrons is given by:

K.E. = hf - \phi

where:

• K.E. is the kinetic energy of the emitted electrons.
• h is Planck's constant.
• f is the frequency of the incident light.
• \phi is the work function of the metal, equal to hv_0, where v_0 is the threshold frequency.

The kinetic energy of emitted electrons can also be expressed in terms of their maximum velocity v as:

\frac{1}{2} mv^2

where m is the electron's mass.

For frequency 2v_0:

The energy equation becomes:

\frac{1}{2} mv_1^2 = h(2v_0) - hv_0 = hv_0

For frequency 5v_0:

The energy equation becomes:

\frac{1}{2} mv_2^2 = h(5v_0) - hv_0 = 4hv_0

Now calculate the ratio of v_1 to v_2:

Using the equations for kinetic energy, we have:

• v_1 = \sqrt{\frac{2hv_0}{m}}
• v_2 = \sqrt{\frac{8hv_0}{m}}

Thus, the ratio \frac{v_1}{v_2} is:

\frac{v_1}{v_2} = \sqrt{\frac{\frac{2hv_0}{m}}{\frac{8hv_0}{m}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Therefore, the ratio of v_1 to v_2 is 1:2.