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  • The maximum kinetic energy of photoelectrons emitted from a certain metallic surface is <img alt="30 \mathrm{eV}" src="https://entrancecorner.oncodecogs.com/gif.latex?30%20%5Cmathrm%7BeV%7D" /

The maximum kinetic energy of photoelectrons emitted from a certain metallic surface is 30 \mathrm{eV}when monochromatic radiation of wavelength \lambda falls on it. When the same surface is illuminated with light of wavelength 2 \lambda, the maximum kinetic energy of photo electrons is observed to be 10 \mathrm{eV}. Then the wavelength \lambda is: \left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{S}=4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)

Option: 1

350 \, \AA


Option: 2

410\, \AA


Option: 3

310 \, \AA


Option: 4

220 \, \AA


Answers (1)

best_answer

Suppose W is the work-function of the metallic surface. Hence in first case, if  \left(E_K^{\max }\right)_1 is

the maximum kinetic energy of the emitted photoelectron, then for the incident radiation of wavelength \lambda, we have

(\mathrm{hc} / \lambda)=\mathrm{W}+\left(\mathrm{E}_{\mathrm{K}}^{\max }\right)_1                                  ------(1)

Similarly, in 2^{\text {nd }} case, if \left(E_K^{\max }\right)_2 is the maximum kinetic energy of the emitted photoelectron, then for the incident light of wavelength 2 \lambda, we have

(\mathrm{hc} / 2 \lambda)=\mathrm{W}+\left(\mathrm{E}_{\mathrm{K}}^{\max }\right)_2                               ------(2)

Substracting eq. (2) from eq.(1), we get

\begin{aligned} & \frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{2 \lambda}=\left(\mathrm{E}_{\mathrm{K}}^{\max }\right)_1-\left(\mathrm{E}_{\mathrm{K}}^{\max }\right)_2 \\ & \text { or } \lambda=\frac{\mathrm{hc}}{2\left[\left(\mathrm{E}_{\mathrm{K}}^{\max }\right)_1-\left(\mathrm{E}_{\mathrm{k}}^{\max }\right)_2\right]} \\ & =\frac{\left(4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{2(30-10) \mathrm{eV}} \\ & =310 \times 10^{-10} \mathrm{~m}=310 \AA \\ & \end{aligned}

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Ritika Jonwal

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