A proton when accelerated through a potential | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda

Vedclass · Accepted Answer

$\frac{V}{8}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For the proton $(p)$: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.For the $\alpha$-particle $(\alpha)$: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}}$.Given $\lambda_p = \lambda_\alpha$, we have $\sqrt{2m_p q_p V} = \sqrt{2m_\alpha q_\alpha V_\alpha}$.Squaring both sides: $m_p q_p V = m_\alpha q_\alpha V_\alpha$.We know that $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.Substituting these values: $m_p q_p V = (4m_p)(2q_p) V_\alpha$.$m_p q_p V = 8 m_p q_p V_\alpha$.Therefore, $V_\alpha = \frac{V}{8}$.

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

Similar Questions

Calculate the de Broglie wavelength of an electron with a kinetic energy of $200 \ eV$. [Mass of electron $= 1 \times 10^{-30} \ kg$,charge on electron $= 1.6 \times 10^{-19} \ C$,Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$].

$A$ proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$.

The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \, nm$ to $0.5 \, nm$ is

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module