The retarding potential necessary to stop the emission of photoelectrons, when a target material of work…

The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function \(1.24 \mathrm{eV}\) is irradiated with light of wavelength \(4.36 \times 10^{-7} \mathrm{~m}\) is
  1. \(4.08 \mathrm{eV}\)
  2. \(2.84 \mathrm{eV}\)
  3. \(1.60 \mathrm{eV}\)
  4. \(0.36 \mathrm{eV}\)

Solution

Given, work function, $\phi_0=1.24 \mathrm{eV}$ $=1.24 \times 1.6 \times 10^{-19} \mathrm{~J}$ Wavelength, $\lambda=4.36 \times 10^{-7} \mathrm{~m}$ According to Einstein's photoelectric equation, $\begin{aligned} e V_0 &= \frac{h c}{\lambda}-\phi_0 \\ \Rightarrow e V_0 &= \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.36 \times 10^{-7}}-1.24 \times 1.6 \times 10^{-19} \\ V_0 &= \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24 \\ &=2.85-1.24=1.61 \mathrm{~V} \simeq 1.60 \mathrm{~V} \end{aligned}$

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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