Question

An electron (mass m) with an initial velocity ${\displaystyle v=v_0\hat{i}(v_0>0)}$ is in an electric field ${\displaystyle E=-E_0\hat{i}}$(E₀ = constant > 0). It’s de Broglie wavelength at time t is given by ______.

विकल्प

• ${\displaystyle \frac{\lambda }{(1+\frac{e{E}_0}{m}\frac{t}{v_0})}}$

• ${\displaystyle {\lambda }_0(1+\frac{e{E}_{0}t}{m{v}_0})}$

• ${\displaystyle {\lambda }_0}$

• ${\displaystyle {\lambda }_{0}t}$

Answer 1

An electron (mass m) with an initial velocity ${\displaystyle v=v_0\hat{i}(v_0>0)}$ is in an electric field ${\displaystyle E=-E_0\hat{i}}$(E₀ = constant > 0). It’s de Broglie wavelength at time t is given by ${\displaystyle \underline{\frac{\lambda }{(1+\frac{e{E}_0}{m}\frac{t}{v_0})}}}$.

Explanation:

Initial de-Broglie wavelength ${\displaystyle {\lambda }_0=\frac{h}{m{v}_0}}$

Force on electron = F = qE v F = (– e)(– E₀i)

${\displaystyle ma=e{E}_{0}i}$

${\displaystyle a=\frac{e{E}_0}{m}i}$

Velocity of electron after time t is v = v₀ + at

${\displaystyle v=v_{0}i+\frac{e{E}_0}{m}i\cdot t}$

${\displaystyle v=[v_0+\frac{e{E}_{0}t}{m}]\hat{i}}$

∴ New de-Broglie wavelength ${\displaystyle \lambda =\frac{h}{mv}}$

${\displaystyle \lambda =\frac{h}{m[v_0+\frac{e{E}_{0}t}{m}]\hat{i}}=\frac{h}{m{v}_0[1+\frac{e{E}_{0}t}{m{v}_0}]}}$

${\displaystyle \lambda =\frac{{\lambda }_0}{[1+\frac{e{E}_{0}t}{m{v}_0}]}}$  ......${\displaystyle [\because \frac{h}{m{v}_0}={\lambda }_0 \text{from equation}I]}$