Question

An electromagnetic wave of wavelength$\lambda $ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Broglie wavelength$\lambda '$ , then,
A. $\lambda =\dfrac{mc}{h}\lambda {{'}^{2}}$
B. $\lambda =\dfrac{3mc}{2h}\lambda {{'}^{2}}$
C. $\lambda =\dfrac{2mc}{h}\lambda {{'}^{2}}$
D. $\lambda =\dfrac{5mc}{h}\lambda {{'}^{2}}$

Answer 1

Hint: As a first step you could recall Einstein’s photoelectric equation. As per the given condition, you could neglect the term representing the work function. Now substitute for the K.E energy in terms of the momentum of the de Broglie wave and then rearrange to get the required expression.

Formula used: Einstein’s photoelectric equation,
$K.{{E}_{\max }}=h\nu -{{\phi }_{0}}$
Expressions for momentum,
$p=mv$
$p=\dfrac{h}{\lambda '}$

Complete step by step answer:
We are given in the question an electromagnetic wave that has a wavelength of$\lambda $. This electromagnetic wave is incident on a photosensitive surface that has negligible work function. We are also said that the emitted photoelectrons have a de Broglie wavelength of$\lambda '$. Then, we are asked to find the correct expression among the given options.
Firstly, let us recall Einstein’s photoelectric equation which is mathematically expressed as,
$K.{{E}_{\max }}=h\nu -{{\phi }_{0}}$ ………………………………. (1)
Where, $K.{{E}_{\max }}$ is the maximum kinetic energy of the emitted photo electrons, h is the Planck’s constant, $\nu $ is the frequency of incident wave and ${{\phi }_{0}}$ is the work function of the surface on which the radiation is incident.
But here the surface is known to have negligible work function and also kinetic energy could be given by,
$K.{{E}_{max}}=\dfrac{1}{2}m{{v}^{2}}$
Also, the frequency can be given by,
$\nu =\dfrac{c}{\lambda }$
$\Rightarrow \dfrac{1}{2}m{{v}^{2}}=h\dfrac{c}{\lambda }$ ……………………………….. (2)
But we know that expression for momentum is given by,
$p=mv$
So, (2) becomes,
$\dfrac{{{p}^{2}}}{2m}=\dfrac{hc}{\lambda }$ ……………………….. (3)
But we know another expression for the momentum of de Broglie wave of wavelength $\lambda '$ as,
$p=\dfrac{h}{\lambda '}$
Substituting in (3) gives,
${{\dfrac{\left( \dfrac{h}{\lambda '} \right)}{2m}}^{2}}=\dfrac{hc}{\lambda }$
Rearranging,
$\lambda =\dfrac{2m\lambda {{'}^{2}}}{{{h}^{2}}}\times hc$
$\therefore \lambda =\dfrac{2mc}{h}\lambda {{'}^{2}}$
Therefore, we found that, the wavelength of incident electromagnetic wave is given by the relation,
$\lambda =\dfrac{2mc}{h}\lambda {{'}^{2}}$

So, the correct answer is “Option C”.

Note: We should know that though the maximum kinetic energy of the emitted photoelectrons varies linearly with the frequency of incident radiation, it doesn’t depend on its intensity. For the frequency of incident radiation which is lower than threshold frequency, no photoelectric emission takes place even for very high intensity. Also, the threshold frequency has different values for different metals.