Question 334: Dual Nature of Matter and Radiation

Question

The light of wavelength '$\lambda$' is incident on the surface of metal of work function $\phi$ and emits the electron. The maximum velocity of electron emitted is [$\mathrm{m}=$ mass of electron and $\mathrm{h}=$ Planck's constant, $\mathrm{c}=$ velocity of light]

Accepted Answer

We have $\frac{1}{2} m v_{\max }^2=\frac{h c}{\lambda}-\phi=\frac{h c-\phi \lambda}{\lambda}$
$\begin{array}{ll}
	herefore \quad \mathrm{V}_{\max }^2=\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda} \\
	herefore \quad\mathrm{V}_{\max }=\sqrt{\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda}}
\end{array}$

Answer

$\left[\frac{2(h c-\lambda)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$

Answer

$\left[\frac{2(h c-\phi) \lambda}{m c}\right]$

Answer

$\left[\frac{2(h c-\lambda)}{m \lambda}\right]$

Answer

$\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$

Dual Nature of Matter and Radiation

The light of wavelength '\lambda' is incident on the surface of metal of work function \phi and emits the electron. The maximum velocity of electron emitted is [\mathrm{m}= mass of electron and \mathrm{h}= Planck's constant, \mathrm{c}= velocity of light]

Choose the correct answer: