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. \(H^{-1/2}\)

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values ${\mathrm{\lambda }}_1, {\mathrm{\lambda }}_2$ with ${\mathrm{\lambda }}_1>{\mathrm{\lambda }}_2$. Which of the following statement/s is/are true?
 

1. (b, d)

2. (a, c)

3. (b, c, d)

4. (a, c, d)

The de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of the electron is ${\mathrm{v}}_{\mathrm{e}}=\frac{\mathrm{c}}{100}$. Then,

1. \(\frac{E_e}{E_p}=10^{-4}\)

2. \(\frac{E_e}{E_p}=10^{-2}\)

3. \(\frac{P_e}{m_ec}=10^{-2}\)

4. \(\frac{P_e}{m_ec}=10^{-4}\)

Two particles A₁ and A₂ of masses \(\mathrm{m_1},\) \(\mathrm{m_2}\) \((​​\mathrm{m_1>m_2})\) have the same de-Broglie wavelength. Then,

1. (b), (c)
2. (a), (c)
3. (c), (d)
4. (b), (d)

Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}{\mathrm{mv}}^2$, becomes comparable with ${\mathrm{mc}}^2$, where m is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?

(a) $\mathrm{\lambda }=10 \mathrm{nm}$
(b) $\mathrm{\lambda }={10}^{-1} \mathrm{nm}$
(c) $\mathrm{\lambda }={10}^{-4} \mathrm{nm}$
(d) $\mathrm{\lambda }={10}^{-6} \mathrm{nm}$

1. (a, c)

2. (a, d)

3. (c, d)

4. (a, b)

An electron (mass m) with an initial velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_0\hat{\mathrm{i}}$ is in an electric field $\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\hat{\mathrm{j}}$. If ${\mathrm{\lambda }}_0=\frac{\mathrm{h}}{{\mathrm{mv}}_0}$, its de-Broglie wavelength at time t is given by:

1. ${\mathrm{\lambda }}_0$

2. ${\mathrm{\lambda }}_0\sqrt{1+\frac{{\mathrm{e}}^2{\mathrm{E}}_0^2{\mathrm{t}}^2}{{\mathrm{m}}^2{\mathrm{v}}_0^2}}$

3. $\frac{{\mathrm{\lambda }}_0}{\sqrt{1+\frac{{\mathrm{e}}^2{\mathrm{E}}_0^2{\mathrm{t}}^2}{{\mathrm{m}}^2{\mathrm{v}}_0^2}}}$

4. $\frac{{\mathrm{\lambda }}_0}{\left(1+\frac{{\mathrm{e}}^2{\mathrm{E}}_0^2{\mathrm{t}}^2}{{\mathrm{m}}^2{\mathrm{v}}_0^2}\right)}$

An electron (mass \(m\)) with an initial velocity \(\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}}\) $\stackrel{}{\mathrm{}}$(\(\mathrm{v}_0>0\)) is in an electric field \(\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{\mathrm{i}}\)$({\mathrm{}}_{}$\(E_o\)=constant\(>0\)). Its de-Broglie wavelength at time \(t\) is given by:

An electron is moving with an initial velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_0\hat{\mathrm{i}}$ and is in a magnetic field $\overrightarrow{\mathrm{B}}={\mathrm{B}}_0\hat{\mathrm{j}}$. Then, its de-Broglie wavelength:

1. remains constant

2. increases with time

3. decreases with time

4. increases and decreases periodically

A proton, a neutron, an electron and an $\mathrm{\alpha }$-particle have the same energy. Then, their de-Broglie wavelengths compare as:

1. ${\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{n}}>{\mathrm{\lambda }}_{\mathrm{e}}>{\mathrm{\lambda }}_{\mathrm{\alpha }}$

2. ${\mathrm{\lambda }}_{\mathrm{\alpha }}<{\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{n}}<{\mathrm{\lambda }}_{\mathrm{e}}$

3. ${\mathrm{\lambda }}_{\mathrm{e}}<{\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{n}}>{\mathrm{\lambda }}_{\mathrm{\alpha }}$

4. ${\mathrm{\lambda }}_{\mathrm{e}}={\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{n}}={\mathrm{\lambda }}_{\mathrm{\alpha }}$