(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_p qV}}$.Substituting the values: $h = 6.63 \times 10^{-

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

(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_p qV}}$.Substituting the values: $h = 6.63 \times 10^{-34} \,J-s$, $m_p = 1.67 \times 10^{-27} \,kg$, $q = 1.6 \times 10^{-19} \,C$, and $V = 1000 \,V$.$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times 1000}}$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{5.344 \times 10^{-43}}} = \frac{6.63 \times 10^{-34}}{7.31 \times 10^{-22}} \approx 9.07 \times 10^{-13} \,m$.Thus, the wavelength is approximately $9.1 \times 10^{-13} \,m$.

Class 12 Physics Dual Nature of Radiation and Matter Matter Waves and de Broglie Wavelength

Easy

If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

TS EAMCET 2018 All TS EAMCET

A
$9.1 \times 10^{-13} \,m$
B
$9.1 \times 10^{13} \,m$
C
$1.09 \times 10^{-15} \,m$
D
$1.09 \times 10^{15} \,m$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Dual Nature of Radiation and Matter

Subtopic

Matter Waves and de Broglie Wavelength

Previous Year

TS EAMCET 2018 Paper All TS EAMCET years →

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

DifficultJEE MAIN 2020

View Solution

$A$ proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelengths is

EasyKCET 2018

View Solution

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:

Medium

View Solution

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)

DifficultJEE MAIN 2024

View Solution

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is equal to :

MediumJEE MAIN 2023

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

Similar Questions

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

$A$ proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelengths is

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:

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)