A proton is accelerated through a potential d | vedclass

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

Question

(D) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For

Vedclass · Accepted Answer

$6.25$

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}}$.For a doubly ionized $^8_4 Be$ nucleus,the mass is $m_{Be} = 8m_p$ and the charge is $q_{Be} = 2q_p$. Thus,$\lambda_{Be} = \frac{h}{\sqrt{2(8m_p)(2q_p)V_{Be}}}$.Given $\lambda_p = \lambda_{Be}$,we equate the denominators:$2m_p q_p V_p = 2(8m_p)(2q_p)V_{Be}$.$m_p q_p V_p = 16 m_p q_p V_{Be}$.$V_p = 16 V_{Be}$.Given $V_p = 100 \ V$,we have $100 = 16 V_{Be}$.$V_{Be} = \frac{100}{16} = 6.25 \ V$.

$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

Similar Questions

For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

Which of the following has the largest de Broglie wavelength?

If a ball of mass $1 \ kg$ has a velocity of $1 \ m/s$,what is its de Broglie wavelength?

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module