Dual Nature of Radiation and Matter - Result Question 22
24. The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} m$ is
(Given $h=4.14 \times 10^{-15} eVs$ and $c=3 \times 10^{8} ms^{-1}$ )
(a) $\approx 0.6 \times 10^{6} ms^{-1}$
(b) $\approx 61 \times 10^{3} ms^{-1}$
(c) $\approx 0.3 \times 10^{6} ms^{-1}$
(d) $\approx 6 \times 10^{5} ms^{-1}$
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Answer:
Correct Answer: 24. (a, d)
Solution:
- (a, d) Both answers are correct
Given,
$\lambda_0=3250 \times 10^{-10} m$
$\lambda=2536 \times 10^{-10} m$
$\phi=\frac{hc}{\lambda_0}=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{3250 \times 10^{-10}}=3.82 eV$
$h v=\frac{h c}{\lambda}=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{2536 \times 10^{-10}}=4.89 eV$
According to Einstein’s photoelectric equation,
$K _{\text{max }}=hv-\phi$
$KE _{\max }=(4.89-3.82) eV=1.077 eV$
$\frac{1}{2} mv^{2}=1.077 \times 1.6 \times 10^{-19}$
$\Rightarrow v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}$
or, $v=0.6 \times 10^{6} m / s$ or $6 \times 10^{5} m / s$
