Question 44: Dual Nature of Matter and Radiation

Question

A photoelectric material having work-function $${\phi _0}$$ is illuminated with light of wavelength $$\lambda \left( {\lambda  < {{he} \over {{\phi _0}}}} \right).$$ The fastest photoelectron has a de-Broglic wavelength $${\lambda _d}.$$ A change in wavelength of the incident light by $$\Delta \lambda $$ result in a change $$\Delta {\lambda _d}$$ in $${\lambda _d}.$$ Then the ratio $$\Delta {\lambda _d}/\Delta \lambda $$ is proportional to

Accepted Answer

Here, $$\lambda$$ is the wavelength of incident light and $$\lambda$$
d
is the de Broglie wavelength of the fastest photoelectron. The fastest ejected photoelectron has the maximum kinetic energy which is given by
$${K_{\max }} = {{hc} \over \lambda } - {\phi _0}$$ .....(1)
The de Broglie wavelength of the photoelectron having kinetic energy $${K_{\max }} = {{{p^2}} \over {2m}}$$ is given by
$${\lambda _d} = {h \over p} = {h \over {\sqrt {2m{K_{\max }}} }}$$ ..... (2)
where m is the mass of the electron and p is its linear momentum. Eliminate K
max
from equations (1) and (2) to get
$${{{h^2}} \over {2m\lambda _d^2}} = {{hc} \over \lambda } - {\phi _0}$$ ...... (3)
Differentiate equation (3) to get
$$( - 2){{{h^2}} \over {2m\lambda _d^3}}\Delta {\lambda _d} = ( - 1){{hc} \over {{\lambda ^2}}}\Delta \lambda $$,
which gives
$${{\Delta {\lambda _d}} \over {\Delta \lambda }} = {{mc} \over h}{{\lambda _d^3} \over {{\lambda ^2}}}$$.

Answer

$${\lambda _d}/\lambda $$

Answer

$$\lambda _d^2/{\lambda ^2}$$

Answer

$$\lambda _d^3/\lambda $$

Answer

$$\lambda _d^3/{\lambda ^2}$$

Dual Nature of Matter and Radiation

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