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7
MCQ (Single Correct Answer)

An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \mathbf{i}\left(v_0>0\right)$ is in an electric field $\mathbf{E}=-E_0 \hat{\mathbf{i}}\left(E_0=\right.$ constant $\left.>0\right)$. It's de-Broglie wavelength at time $t$ is given by

A
$\frac{\lambda_0}{\left(1+\frac{e E_0}{m} \frac{t}{v_0}\right)}$
B
$\lambda_0\left(1+\frac{e E_0 t}{m v_0}\right)$
C
$\lambda_0$
D
$\lambda_0 t$
8
MCQ (Single Correct Answer)

An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ is in an electric field $\mathbf{E}=E_0 \hat{\mathbf{j}}$. If $\lambda_0=h / m v_0$, it's de-Broglie wavelength at time $t$ is given by

A
$\lambda_0$
B
$\lambda_0 \sqrt{1+\frac{\mathrm{e}^2 E_0^2 t^2}{m^2 v_0^2}}$
C
$\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$
D
$\frac{\lambda_0}{\left(1+\frac{\mathrm{e}^2 E_0^2 t^2}{m^2 v_0^2}\right)}$
9
MCQ (Multiple Correct Answer)

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

A
$\lambda=10 \mathrm{~nm}$
B
$\lambda=10^{-1} \mathrm{~nm}$
C
$\lambda=10^{-4} \mathrm{~nm}$
D
$\lambda=10^{-6} \mathrm{~nm}$
10
MCQ (Multiple Correct Answer)

Two particles $A_1$ and $A_2$ of masses $m_1, m_2\left(m_1>m_2\right)$ have the same de-Broglie wavelength. Then,

A
their momenta are the same
B
their energies are the same
C
energy of $A_1$ is less than the energy of $A_2$
D
energy of $A_1$ is more than the energy of $A_2$
11
MCQ (Multiple Correct Answer)

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

A
$\frac{E_e}{E_p}=10^{-4}$
B
$\frac{E_e}{E_p}=10^{-2}$
C
$\frac{p_e}{m_e c}=10^{-2}$
D
$\frac{p_e}{m_e C}=10^{-4}$
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