Modern Physics Ques 92
- In a photoelectric effect experiment, the threshold wavelength of light is $380 $ $nm$. If the wavelength of incident light is $260$ $ nm$, the maximum kinetic energy of emitted electrons will be
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Answer:
Correct Answer: 92.(c)
Solution:
Formula:
- Given, threshold wavelength, $\lambda _0=380$ $ nm$
Wavelength of incident light, $\lambda=260 $ $nm$
Using Einstein’s relation of photoelectric effect,
$(KE) _{\max } =e V _0=h \nu-h v _0 \quad …….(i)$
$\text { But } h \nu =E=\frac{1237}{\lambda(nm)} eV $
$\therefore E _0 =\frac{1237}{\lambda _0(nm)} eV \quad …….(ii)$
From Eqs. (i) and (ii), we get
$ \begin{aligned} (KE) _{\max } & =E-E _0 \left(\frac{1237}{\lambda}-\frac{1237}{\lambda _0}\right) eV \\ & =1237 \left[\frac{1}{\lambda}-\frac{1}{\lambda _0}\right] eV(\lambda \text { in nm }) \quad …….(iii) \end{aligned} $
By putting values of $\lambda$ and $\lambda _0$ in Eq. (iii), we get
$ \begin{aligned} (KE) _{\max } & =1237 \left(\frac{1}{260}-\frac{1}{380}\right) eV \\ & =1237 \times \left[\frac{380-260}{380 \times 260}\right] eV \\ \Rightarrow(KE) _{\max } & =1.5 eV \end{aligned} $
BETA
