Modern Physics 2 Question 23
22. When photons of energy $4.25 eV$ strike the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T _A$ expressed in $eV$ and de-Broglie wavelength $\lambda _A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70 eV$ is $T _B=\left(T _A-1.50 eV\right)$. If the de-Broglie wavelength of these photoelectrons is $\lambda _B=2 \lambda _A$, then
$(1994,2 M)$
(a) the work function of $A$ is $2.25 eV$
(b) the work function of $B$ is $4.20 eV$
(c) $T _A=2.00 eV$
(d) $T _B=2.75 eV$
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Solution:
- $K _{\max }=E-W$
Therefore,
$$ \begin{aligned} T _A & =4.25-W _A \\ T _B & =\left(T _A-1.50\right)=4.70-W _B \end{aligned} $$
From Eqs. (i) and (ii),
$$ W _B-W _A=1.95 eV $$
de-Broglie wavelength is given by
$$ \lambda=\frac{h}{\sqrt{2 K m}} \quad \text { or } \quad \lambda \propto \frac{1}{\sqrt{K}} \quad K=KE \text { of electron } $$
$$ \therefore \quad \frac{\lambda _B}{\lambda _A}=\sqrt{\frac{K _A}{K _B}} \text { or } 2=\sqrt{\frac{T _A}{T _A-1.5}} $$
This gives, $\quad T _A=2 eV$
From Eq. (i) $W _A=4.25-T _A=2.25 eV$
From Eq. (iii) $W _B=W _A+1.95 eV=(2.25+1.95) eV$
or
$$ \begin{aligned} W _B & =4.20 eV \\ T _B & =4.70-W _B=4.70-4.20 \\ & =0.50 eV \end{aligned} $$
