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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its speed

Vedclass · Accepted Answer

$\lambda_1 > \lambda_2$ or $\lambda_1 < \lambda_2$

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its speed.When an electron enters a uniform electric field,the electric force $F = qE$ acts on it,which can either accelerate or decelerate the electron depending on the direction of the field. Thus,the speed $v_1$ changes,causing $\lambda_1$ to change.When an electron enters a uniform magnetic field,the magnetic force $F = q(v 	imes B)$ acts perpendicular to the velocity. This force changes the direction of motion but does not change the magnitude of the speed. Thus,the speed $v_2$ remains constant,and $\lambda_2$ remains constant.Since the speed $v_1$ in the electric field can either increase or decrease relative to the initial speed,$\lambda_1$ can be either smaller or larger than $\lambda_2$.

Two electrons are moving with the same speed $v$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. After some time,if the de-Broglie wavelengths of the two are $\lambda_1$ and $\lambda_2$,then:

Similar Questions

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression:

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

Photons of wavelength $\lambda$ are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons is (work function is negligible).
($c =$ velocity of light,$h =$ Planck's constant,$m =$ mass of electron)

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