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:

  1. (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.