Question

In a photocell, with excitation wavelength $\lambda $ , the faster electron has speed v. If the excitation wavelength is changed to $\dfrac{{3\lambda }}{4}$ , the speed of the fastest electron will be
(A) ${V^{{{\left( {\dfrac{3}{4}} \right)}^{\dfrac{1}{2}}}}}$
(B) ${V^{{{\left( {\dfrac{4}{3}} \right)}^{\dfrac{1}{2}}}}}$
(C) less than ${V^{{{\left( {\dfrac{4}{3}} \right)}^{\dfrac{1}{2}}}}}$
(D) greater than ${V^{{{\left( {\dfrac{4}{3}} \right)}^{\dfrac{1}{2}}}}}$

Answer 1

Hint: Hint- Use Einstein’s photoelectric emission based on Planck's quantum theory of light and express the formula in terms of $\lambda $ . According to the photoelectric equation, $V \propto \dfrac{1}{\lambda }$ . Thus, as the value of wavelength decreases, the speed of the fastest electron increases.

Complete step by step solution:
Let the wavelength of the incident excitation be denoted by $\lambda $ .
Let the speed of the fastest electron be denoted by V .
Let the work function be denoted by $\phi $ .
We know that,
When light of certain frequency (greater than work function) is incident on a metallic surface, the energy of the incident photon is partly used to eject electrons from surfaces and partly remains as kinetic energy of ejected electrons. This phenomenon was formulated by Einstein and was given as:
Energy of photon = Work Function + Kinetic Energy of electron; i.e
$\dfrac{{hc}}{\lambda } = \phi + \dfrac{1}{2}m{v^2}.....(1)$
where $h$ = Planck’s constant
$m$= mass of an electron
When the excitation wavelength is changed to $\dfrac{{3\lambda }}{4}$ , let the new velocity of the fastest electron be denoted by V.
The photoelectric equation now becomes:
$\dfrac{{4hc}}{{3\lambda }} = \phi + \dfrac{1}{2}m{V^2}........(2)$
Multiplying equation (1) by a factor of $\dfrac{4}{3}$ gives:
$\dfrac{{4hc}}{{3\lambda }} = \dfrac{{4\phi }}{3} + \dfrac{4}{3}\left( {\dfrac{1}{2}m{v^2}} \right).......(3)$
Subtracting equation (2) from equation (3) we get
$\dfrac{{4hc}}{{3\lambda }} - \dfrac{{4hc}}{{3\lambda }} = \dfrac{{4\phi }}{3} - \phi + \left( {\dfrac{4}{3}\left( {\dfrac{1}{2}m{v^2}} \right) - \dfrac{1}{2}m{V^2}} \right)$
$ \Rightarrow 0 = \dfrac{\phi }{3} + \left( {\dfrac{4}{3}\left( {\dfrac{1}{2}m{v^2}} \right) - \dfrac{1}{2}m{V^2}} \right)$
We know that the work function is constant for a given material. Thus we can rewrite the equation as follows:
$\dfrac{1}{2}m{V^2} = \dfrac{\phi }{3} + \dfrac{4}{3}\left( {\dfrac{1}{2}m{v^2}} \right)$
$ \Rightarrow {V^2} = \dfrac{{2\phi }}{{3m}} + \dfrac{{4{v^2}}}{3}$
$ \Rightarrow {V^2} = K + \dfrac{{4{v^2}}}{3}$
where $K$ is a constant quantity, as $\phi $ and $m$ are both constants.
Therefore,
${V^2} > \dfrac{{4{v^2}}}{3}$
$ \Rightarrow V > \sqrt {\dfrac{4}{3}} v$

Hence the Option (D) is correct.

Note: Please note that we are establishing a quantitative relation between the before and after speeds of the fastest electrons, therefore do not commit the mistake of subtracting the first two equations immediately. Also, please notice that as the wavelength of the excitation is decreased, the K.E of the fastest electron should increase, thus increasing the speed too. Increase in intensity of the light increases the number of photoelectrons, not the kinetic energy.