An electron of mass $m$ is moving in an electric field $\vec{E}=-2 E_{\mathrm{o}} \hat{i}\left(E_{\mathrm{o}}=\right.$ constant $\left.>0\right)$, with an initial velocity $\vec{V}=v_{\mathrm{o}} \hat{i} \left(v_{\mathrm{o}}=\right.$ constant $\left.>0\right)$. If $\lambda_{\mathrm{o}}=\frac{h}{4 m v_{\mathrm{o}}}$, its de Broglie wavelength at time $t$ is

$\_\_\_\_$ .

( $e=$ charge of electron)

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 $\alpha \times 10^{-19} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .

$$ \left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} \text { and } \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) $$

The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_{\mathrm{e}}$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_{\mathrm{p}}$. If their corresponding masses are $m_{\mathrm{e}}$ and $m_{\mathrm{p}}$, respectively, then the ratio of their de Broglie wavelengths $\left(\frac{\lambda_e}{\lambda_p}\right)$ is $\_\_\_\_$ .

The graph shows variation of stopping potential $V_{\mathrm{o}}$ with the frequency $v$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?