Question 327: Dual Nature of Matter and Radiation

Question

When radiations of wavelength $$\lambda$$ is incident on a metallic surface the stopping potential required is $$4.8 \mathrm{~V}$$. If same surface is illuminated with radiations of double the wavelength, then required stopping potential becomes $$1.6 \mathrm{~V}$$, then the value of threshold wavelength for the surface is

Accepted Answer

From Einstein's photoelectric equation,
$$\begin{array}{rlrl} 
& \mathrm{KE}_{\max } & =h v-h v_0 \
	ext { or } & e V & =h v-h v_0 \
\Rightarrow \quad & \quad e V & =\frac{h c}{\lambda}-\frac{h c}{\lambda_0}
\end{array}$$
Here, $$V$$ is stopping potential and $$\lambda$$ is the incident wavelength, whereas $$\lambda_0$$ its threshold wavelength.
For case $$1 e(4.8)=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}$$ ..... (i)
For case $$2 e(1.6)=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0}$$ ..... (ii)
From Eqs. (i) and (ii), we get
$$\begin{array}{ll} 
& e 	imes(3.2)=\frac{h c}{2 \lambda} \
\Rightarrow \quad & e(1.6)=\frac{h c}{4 \lambda} \quad 	ext{.... (iii)}
\end{array}$$
Putting the Eq. (iii) in Eq. (ii), we get
$$\begin{aligned}
\Rightarrow \quad & \frac{h c}{4 \lambda}=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0} \
\Rightarrow \quad & \frac{h c}{\lambda_0}=\frac{h c}{2 \lambda}-\frac{h c}{4 \lambda} \
\Rightarrow \quad & \frac{h c}{\lambda_0}=\frac{h c}{4 \lambda} \
\Rightarrow \quad & \lambda_0=4 \lambda
\end{aligned}$$

Answer

$$2 \lambda$$

Answer

$$4 \lambda$$

Answer

$$6 \lambda$$

Answer

$$8 \lambda$$

Dual Nature of Matter and Radiation

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