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  • When a surface is irradiated with light of wavelength , a photo current appears which vanis

When a surface is irradiated with light of wavelength 4950 \, \AA, a photo current appears which vanishes if a retarding potential equal to or greater than 0.6 volt is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Then the wavelength of the second source.
 

Option: 1

411 \, \AA


Option: 2

4125 \, \AA


Option: 3

1025 \, \AA


Option: 4

2245 \, \AA


Answers (1)

best_answer

According to Einstein’s photoelectric equation:

(h c / \lambda)=W+E_k^{\max }=W+e V_0

where V_0 is the stopping potential and W is the work function of the emitting surface. Hence in 1^{\text {st }} case,

\left(\mathrm{hc} / \lambda_1\right)=\mathrm{W}+\mathrm{e} \mathrm{V}_1 or \mathrm{W}=\left(\mathrm{hc} / \lambda_1\right)-\mathrm{eV}{ }_1

=\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{4950 \times 10^{-10}}-\left(1.6 \times 10^{-19}\right) \times 0.6

=4.0 \times 10^{-19}-0.96 \times 10^{-19}

=3.04 \times 10^{-19} \mathrm{~J}

=\frac{3.04 \times 10^{-19}}{1.6 \times 10^{-19}}=1.9 \mathrm{eV}

Using the light of another wavelength \lambda_2, we shall have 

\left(\mathrm{hc} / \lambda_2\right)=\mathrm{W}+\mathrm{eV}_2=3.04 \times 10^{-19}+\left(1.6 \times 10^{-19}\right) \times 1.1 \\=4.80 \times 10^{-19} \text { Joule }

\therefore \lambda_2=\frac{\text { hc }}{4.80 \times 10^{-19}}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{4.80 \times 10^{-19}}

=4125 \times 10^{-10} \mathrm{~m}=4125 \AA

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Riya

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