Dual Nature of Radiation and Matter - Result Question 43
45. The work function of a surface of a photosensitive material is $6.2 eV$. The wavelength of incident radiation for which the stopping potential is $5 V$ lies in the:
[2008]
(a) Ultraviolet region
(b) Visible region
(c) Infrared region
(d) X-ray region
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Answer:
Correct Answer: 45. (a)
Solution:
- (a) Work function, $\phi_0=6.2 eV$.
Stopping potential $V_0=5 V$.
As, $eV_0=h v-\phi_0$
or $eV_0+\phi_0=\frac{hc}{l}$
or $1=\frac{hc}{eV_0+f_0}$
$\lambda=\frac{6.62 \times 10^{-34} Js \times 3 \times 10^{8} ms^{-1}}{1.6 \times 10^{-19} \times 5 J+6.2 \times 1.6 \times 10^{-19} J}$
$=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}(5+6.2)} m$
$=\frac{19.86 \times 10^{-26}}{17.92 \times 10^{-19}}=1.10 \times 10^{-7} m$.
Thus the wavelength of the incident radiation lies in the ultraviolet region.
