Modern Physics 2 Question 3
3. 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: 3. (c)
Solution:
- Given, threshold wavelength, $\lambda _0=380 nm$
Wavelength of incident light, $\lambda=260 nm$
Using Einstein’s relation of photoelectric effect,
$$ \begin{aligned} & (KE) _{\max } & =e V _0=h \nu-h v _0 \\ \text { But } & h \nu & =E=\frac{1237}{\lambda(nm)} eV \\ \therefore & E _0 & =\frac{1237}{\lambda _0(nm)} eV \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ \begin{aligned} (KE) _{\max } & =E-E _0 \frac{1237}{\lambda}-\frac{1237}{\lambda _0} eV \\ & =1237 \frac{1}{\lambda}-\frac{1}{\lambda _0} eV(\lambda \text { in nm }) \end{aligned} $$
By putting values of $\lambda$ and $\lambda _0$ in Eq. (iii), we get
$$ \begin{aligned} (KE) _{\max } & =1237 \frac{1}{260}-\frac{1}{380} eV \\ & =1237 \times \frac{380-260}{380 \times 260} eV \\ \Rightarrow(KE) _{\max } & =1.5 eV \end{aligned} $$
