Question 22: Dual Nature of Matter and Radiation

Question

Let $$K_1$$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $$\lambda_1$$ and $$K_2$$ corresponding to wavelength $$\lambda_2$$. If $$\lambda_1=2 \lambda_2$$, then

Accepted Answer

Here, $${K_1} = {{hc} \over {{\lambda _1}}} - W$$ .... (i)
and $${K_2} = {{hc} \over {{\lambda _2}}} - W$$ .... (ii)
Substituting $$\lambda_1=2\lambda_2$$ in Eq. (i), we get
$${K_1} = {{hc} \over {2{\lambda _2}}} - W$$
$$ \Rightarrow {K_1} = {1 \over 2}\left( {{{hc} \over {{\lambda _2}}}} \right) - W = {1 \over 2}({K_2} + W) - W$$
$${K_1} = {{{K_2}} \over 2} - {W \over 2}$$
$$ \Rightarrow {K_1} < {{{K_2}} \over 2}$$

Answer

$$2 K_1=K_2$$

Answer

$$K_1=2 K_2$$

Answer

$$K_1 < K_2 / 2$$

Answer

$$K_1>2 K_2$$

Dual Nature of Matter and Radiation

Let $K_1 be the maximum kinetic energy of photoelectrons emitted by light of wavelength \lambda_1 and K_2 corresponding to wavelength \lambda_2. If \lambda_1=2 \lambda_2$, then

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