Question 34: Dual Nature of Matter and Radiation

Question

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V . The work function of the used metal in the experiment is $\alpha 	imes 10^{-19} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .
$$ \left(\mathrm{h}=6.62 	imes 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{e}=1.6 	imes 10^{-19} \mathrm{C} 	ext { and } \mathrm{c}=3 	imes 10^8 \mathrm{~m} / \mathrm{s}ight) $$

Accepted Answer

Use EinsteinÃ¢ÂÂs photoelectric equation:
$$ h
u=\phi+K_{\max} $$
Here,
$\phi$ = work function
$K_{\max}=eV_0$
$V_0=0.2\ 	ext{V}$
Also,
$$ 
u=\frac{c}{\lambda} $$
So,
$$ \phi=\frac{hc}{\lambda}-eV_0 $$
Now substitute the values:
$$ h=6.62	imes 10^{-34}\ 	ext{J s},\quad c=3	imes 10^8\ 	ext{m/s},\quad \lambda=331\ 	ext{nm}=331	imes 10^{-9}\ 	ext{m} $$
First, calculate photon energy:
$$ \frac{hc}{\lambda} = \frac{6.62	imes 10^{-34}	imes 3	imes 10^8}{331	imes 10^{-9}} $$
$$ = \frac{19.86	imes 10^{-26}}{331	imes 10^{-9}} = \frac{19.86}{331}	imes 10^{-17} $$
$$ \approx 0.06	imes 10^{-17} =6	imes 10^{-19}\ 	ext{J} $$
Now,
$$ eV_0=(1.6	imes 10^{-19})(0.2)=0.32	imes 10^{-19}\ 	ext{J} $$
Therefore,
$$ \phi=6	imes 10^{-19}-0.32	imes 10^{-19} $$
$$ \phi=5.68	imes 10^{-19}\ 	ext{J} $$
Comparing with $\alpha	imes 10^{-19}\ 	ext{J}$,
$$ \alpha=5.68 $$
Therefore, the correct answer is:
$$ \boxed{5.68} $$
Option C.

Answer

3.68

Answer

4.68

Answer

5.68

Answer

2.68

Dual Nature of Matter and Radiation

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