A particle is dropped from a height $H.$ The de-Broglie wavelength of the particle as a function of height is proportional to:
1. $H$
2. $H^{1/2}$
3. $H^{0}$
4. $H^{-1/2}$

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1~\text{MeV}$ energy is nearly:
1. $1.2~\text{nm}$
2. $1.2\times 10^{-3}~\text{nm}$
3. $1.2\times 10^{-6}~\text{nm}$
4. $1.2\times 10~\text{nm}$

Consider a beam of electrons (each electron with energy $E_0$) incident on a metal surface kept in an evacuated chamber. Then:

 
Choose the correct option from the given ones:

 
Choose the correct option:

The de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of the electron is $v_e = \dfrac c {100}.$ Then:

1. $\dfrac{E_e}{E_p}=10^{-4}$

2. $\dfrac{E_e}{E_p}=10^{-2}$

3. $\dfrac{P_e}{m_ec}=10^{-2}$

4. $\dfrac{P_e}{m_ec}=10^{-4}$

Choose the correct option:
1. (a), (c)
2. (a), (d)
3. (c), (d)
4. (a), (b)

An electron (mass $m$) with an initial velocity $\overset{\rightarrow}{v} = v_{0} \hat{i}$ is in an electric field $\overset{\rightarrow}{E} = E_{0} \hat{j}$. If $\lambda_{0} = \dfrac{h}{ {mv}_0}$, its de-Broglie wavelength at time $t$ is given by:

1. $\lambda_0$

2. $\lambda_{0} \sqrt{1 + \dfrac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}$

3. $\dfrac{\lambda_{0}}{\sqrt{1 + \dfrac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

4. $\dfrac{\lambda_{0}}{\left(1 + \dfrac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}\right)}$