Question

A photosensitive metallic surface has a work function $$h{\nu _0}$$ . If the photons of energy $2h{\nu _0}$ fall on this surface the electrons come out with a maximum velocity of $4 \times {10^6}m{s^{ - 1}}$ . When the photon energy is increases to $5h{\nu _0}$ then the maximum velocity of photo electrons will be
A. $2 \times {10^6}\,m{s^{ - 1}}$
B. $2 \times {10^7}\,m{s^{ - 1}}$
C. $8 \times {10^5}\,m{s^{ - 1}}$
D. $8 \times {10^6}\,m{s^{ - 1}}$

Answer 1

Hint: This is the concept of photoelectric effect in which photons are incident on the metallic surface and the electrons are emitted from this surface. Here, we will use Einstein's photoelectric equation to calculate the velocity of the emitted electrons. We will equate Einstein's photoelectric equation in case of $2h{v_0}$ and $5h{v_0}$ so that we can get the value of velocity of electrons.

Formula used:
The Einstein’s photoelectric equation is given by
$K.E. = hv - W$
Here, $K.E.$ is the kinetic energy of the emitted electrons, $h$ is the Planck’s constant, $v$ is the velocity of the photons and $W$ is the work done.
Now, the work done according to Einstein’s quantum mechanics is given by
$W = h{v_0}$
Here, $W$ is the work done, $h$ is the Planck’s constant and ${v_0}$ is the velocity of the emitted electrons.

Complete step by step answer:
Consider a photosensitive metallic surface having work function $h{\nu _0}$ .
Now, the Einstein’s photo-electric equation is given by
$K.E. = hv - W$
Now, according to Einstein’s quantum mechanics, the work done is given by
$W = h{v_0}$
Now, putting this value in the above equation, we get
$K.E. = hv - h{v_0}$
Now, when the photons of energy $2h{\nu _0}$ fall on this surface, the electrons from this surface will come out with a maximum velocity of $4 \times {10^6}m{s^{ - 1}}$ , then the Einstein’s photo-electric equation is given by
$\dfrac{1}{2}mv_{\max }^2 = 2h{v_0} - h{v_0}$
$ \Rightarrow \,\dfrac{1}{2}m \times \left( {4 \times {{10}^6}} \right) = h{v_0}$
Now, when the energy of the photon is increased to $5h{v_0}$ , then the Einstein’s photo-electric equation is given by
$\dfrac{1}{2}mv_{\max }^2 = 5h{v_0} - h{v_0}$
$ \Rightarrow \,\dfrac{1}{2}mv_{\max }^2 = 4h{v_0}$
Now, putting the value of $h{v_0}$ , we get
$\dfrac{1}{2}mv_{\max }^2 = 4 \times \dfrac{1}{2}m \times {\left( {4 \times {{10}^6}} \right)^2}$
$ \Rightarrow \,v_{\max }^2 = 64 \times {10^{12}}$
$ \therefore \,{v_{\max }} = 8 \times {10^6}\,m{s^{ - 1}}$
Therefore, when the photon energy is increased to $5h{\nu _0}$ then the maximum velocity of photo electrons will be $8 \times {10^6}\,m{s^{ - 1}}$ .

Hence, option D is the correct option.

Note:An alternate way to solve the above question is given by
$E = W + \dfrac{1}{2}mv_{\max }^2$
Now, the energy of photons is $2h{v_0}$ , therefore, the above equation will become
$2h{v_0} = h{v_0} + \dfrac{1}{2}mv_1^2$
$ \Rightarrow \,h{v_0} = \dfrac{1}{2}mv_1^2$
Similarly, the Einstein’s photoelectric equation when the energy of the photons will increase to $5h{v_0}$ is given by
$ \Rightarrow \,4h{v_0} = \dfrac{1}{2}mv_2^2$
Now, dividing both the equations, we get
$\dfrac{4}{1} = {\left( {\dfrac{{{v_2}}}{{{v_1}}}} \right)^2}$
$ \Rightarrow \,\dfrac{{{v_2}}}{{{v_1}}} = 2$
$ \Rightarrow \,{v_2} = 2{v_1}$
$ \Rightarrow \,{v_2} = 2 \times 4 \times {10^6}$
$ \therefore \,{v_2} = 8 \times {10^6}\,m{s^{ - 1}}$
This is the required answer.