Atomic Structure 1 Question 3
3. What is the work function of the metal, if the light of wavelength $4000 \AA$ generates photoelectron of velocity $6 \times 10^{5} \mathrm{~ms}^{-1}$ from it?
(Mass of electron $=9 \times 10^{-31} \mathrm{~kg}$
Velocity of light $=3 \times 10^{8} \mathrm{~ms}^{-1}$
Planck’s constant $=6.626 \times 10^{-34} \mathrm{Js}$
Charge of electron $=1.6 \times 10^{-19} \mathrm{JeV}^{-1}$ )
(2019 Main, 12 Jan I)
(a) $4.0 \mathrm{eV}$
(b) $2.1 \mathrm{eV}$
(c) $0.9 \mathrm{eV}$
(d) $3.1 \mathrm{eV}$
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Answer:
Correct Answer: 3. (d)
Solution:
- Work function of metal $(\phi)=h v_{0}$ where, $v_{0}=$ threshold frequency
Also, $\quad \frac{1}{2} m_{e} v^{2}=h v-h v_{0}$
or
$$ \begin{aligned} & \frac{1}{2} m_{e} v^{2}=h v-\phi \ & \frac{1}{2} m_{e} v^{2}=\frac{h c}{\lambda}-\phi \end{aligned} $$
Given : $\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}$
$$ \begin{aligned} v & =6 \times 10^{5} \mathrm{~ms}^{-1}, \ m_{e} & =9 \times 10^{-31} \mathrm{~kg}, c=3 \times 10^{8} \mathrm{~ms}^{-1} \ h & =6.626 \times 10^{-34} \mathrm{Js} \end{aligned} $$
Thus, on substituting all the given values in Eq. (i), we get
$$ \begin{aligned} & \frac{1}{2} \times 9 \times 10^{-31} \mathrm{~kg} \times\left(6 \times 10^{5} \mathrm{~ms}^{-1}\right)^{2} \ &=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{4000 \times 10^{-10} \mathrm{~m}}-\phi \ & \therefore \quad \phi=1.62 \times 10^{-21} \mathrm{kgm}^{2} \mathrm{~s}^{-2}-4.96 \times 10^{-19} \mathrm{~J} \ &=3.36 \times 10^{-19} \mathrm{~J} \quad\left[1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}=1 \mathrm{~J}\right] \ &=2.1 \mathrm{eV} \end{aligned} $$
