If lambda_0 is the de-Broglie wavelength for | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{\lambda_0}{2 \sqrt{2}}$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_0 = \frac{h}{\sqrt{2m_p q_p V}}$.For an $\alpha$-particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2q_p$.Substituting these into the formula for the $\alpha$-particle:$\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2q_p)V}} = \frac{h}{\sqrt{8(2m_p q_p V)}} = \frac{1}{\sqrt{8}} \left( \frac{h}{\sqrt{2m_p q_p V}} \right)$.Since $\sqrt{8} = 2\sqrt{2}$,we get $\lambda_{\alpha} = \frac{\lambda_0}{2\sqrt{2}}$.

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

Similar Questions

If $\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} dt=\frac{1}{2}[g(t)]^2+c$,(where $c$ is a constant of integration),then $g(2)$ is

Identify $A, B$ and $C$ in the following reactions:
$CH_3 Cl$ $\xrightarrow{KCN} A$ $\xrightarrow{H_3 O^{\oplus}} B$ $\xrightarrow{C_2 H_5 OH / H^{+}, \Delta} C$

Which of the following oxides is different from the other three oxides?

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Given that the slope of the tangent to a curve $y = y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2 + y^2 - 2x - 2y = 0$,then its equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module