Question 25: Dual Nature of Matter and Radiation

Question

If K$$_1$$ and K$$_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelengths $$\lambda_1$$ and $$\lambda_2$$, respectively incident on a metallic surface and $$\lambda_1=3\lambda_2$$, then

Accepted Answer

Let $$W_0$$ be the work function of the metal, then kinetic energy of photoelectrons emitted when light of wavelength $$\lambda_1$$ is incident will be
$$K_1=\frac{h c}{\lambda_1}-W_0$$
Similarly, $$\quad K_2=\frac{h c}{\lambda_2}-W_0$$
$$\begin{aligned}	ext { Now, } K_1-K_2 & =\frac{h c}{\lambda_1}-\frac{h c}{\lambda_2}=h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}ight] \
& =h c\left[\frac{1}{3 \lambda_2}-\frac{1}{\lambda_2}ight] \
K_1-K_2 & =\frac{-2 h c}{3 \lambda_2}=\frac{-2}{3}\left(K_2+W_0ight) \
\Rightarrow \quad K_1 & =K_2-\frac{2}{3} K_2-\frac{2}{3} W_0 \
& =\frac{K_2}{3}-\frac{2}{3} W_0 \Rightarrow K_1<\frac{K_2}{3}\end{aligned}$$

Answer

$${K_1} > \left( {{{{K_2}} \over 3}} \right)$$

Answer

$${K_1} < \left( {{{{K_2}} \over 3}} \right)$$

Answer

$${K_1} = 2{K_2}$$

Answer

$${K_2} = 2{K_1}$$

Dual Nature of Matter and Radiation

If K$_1 and K_2 are maximum kinetic energies of photoelectrons emitted when lights of wavelengths \lambda_1 and \lambda_2, respectively incident on a metallic surface and \lambda_1=3\lambda_2$, then

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