Atomic Structure - Result Question 27
####3. What is the work function of the metal, if the light of wavelength $4000 \AA$ generates photoelectron of velocity $6 \times 10^{5} ms^{-1}$ from it?
(Mass of electron $=9 \times 10^{-31} kg$
Velocity of light $=3 \times 10^{8} ms^{-1}$
Planck’s constant $=6.626 \times 10^{-34} Js$
Charge of electron $=1.6 \times 10^{-19} JeV^{-1}$ )
(2019 Main, 12 Jan I)
(a) $4.0 eV$
(b) $2.1 eV$
(c) $0.9 eV$
(d) $3.1 eV$
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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 : $\lambda=4000 \AA=4000 \times 10^{-10} m$
$$ \begin{aligned} v & =6 \times 10^{5} ms^{-1}, \ m _e & =9 \times 10^{-31} kg, c=3 \times 10^{8} ms^{-1} \ h & =6.626 \times 10^{-34} Js \end{aligned} $$
Thus, on substituting all the given values in Eq. (i), we get
$$ \begin{aligned} & \frac{1}{2} \times 9 \times 10^{-31} kg \times\left(6 \times 10^{5} ms^{-1}\right)^{2} \ &=\frac{6.626 \times 10^{-34} J s \times 3 \times 10^{8} ms^{-1}}{4000 \times 10^{-10} m}-\phi \ & \therefore \quad \phi=1.62 \times 10^{-21} kgm^{2} s^{-2}-4.96 \times 10^{-19} J \ &=3.36 \times 10^{-19} J \quad\left[1 kg m^{2} s^{-2}=1 J\right] \ &=2.1 eV \end{aligned} $$
