Atomic Structure - Result Question 3
3. What is the work function of the metal, if the light of wavelength $4000 $ Å generates photoelectron of velocity $6 \times 10^5 \mathrm{~ms}^{-1}$ 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. (b)
Solution:
- Work function of metal $(\phi)=h v _0$
where, $v _0=$ threshold frequency
Also, $\quad \dfrac{1}{2} m _e v^{2}=h v-h v _0$
or
$ \begin{aligned} & \dfrac{1}{2} m _e v^{2}=h v-\phi \hspace{10mm} …(i) \\ & \dfrac{1}{2} m _e v^{2}=\dfrac{h c}{\lambda}-\phi \hspace{10mm}…(ii) \end{aligned} $
Given : $\lambda=4000 $ $Å=4000 \times 10^{-10} m$
$ \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
$ \dfrac{1}{2} \times 9 \times 10^{-31} kg \times\left(6 \times 10^{5} ms^{-1}\right)^{2} $
$=\dfrac{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 \hspace{30mm}\left[1 kg m^{2} s^{-2}=1 J\right] $
$=2.1 eV$
BETA
