The de-Broglie wavelength of a proton (charge =1.6×10^(-19) C,m=1.6×10^(-27) kg) accelerated through a potential difference of 1 kV is

Correct answer: (B) 0.9×10^(-12) m. de-Broglie wavelength, λ=h/p=h√2mE=h√2mqV ⇒ λ=(6.6×10^(-34)√(2×1.6×10^(-27)×1.6×10^(-19)×1000)⇒ λ=6.6×10^(-34))/(7.16×10^(-22))=0.9×10^(-12) m

Q99 / 125 · BITSAT 2017 · Physics

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{- 19} C, m = 1.6 \times 10^{- 27} kg$ ) accelerated through a potential difference of $1 kV$ is

A
$600 Å$
Correct answer
B
$0.9 \times 10^{- 12} m$
C
$7 Å$
D
$0.9 nm$

Solution

de-Broglie wavelength,

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m q V}}$

$\Rightarrow λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.6 \times 10^{- 27} \times 1.6 \times 10^{- 19} \times 1000}}$

$\Rightarrow λ = \frac{6.6 \times 10^{- 34}}{7.16 \times 10^{- 22}} = 0.9 \times 10^{- 12} m$

Hence, the correct answer is option: 'B'.

Relevant Questions

Proton (p) and electron (e) will have same deBroglie wavelength when the ratio of their momentum is (assume, )2023 (Memory Based Paper 1)Matter Waves
A proton and an -particle are accelerated from rest by and potentials, respectively. The ratio of their de-Broglie wavelength is:2023 (Memory Based Paper 2)Matter Waves
Electrons are accelerated through a potential difference and protons are accelerated through a potential difference . The de - Broglie wavel2021Matter Waves
A particle X of mass m and initial velocity u collides with another particle Y of mass 3 m 4 , which is at rest. The collision is head-on an2019Matter Waves
An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of . The kinetic energy and the de-Broglie wavelength of2011Matter Waves
A material particle with a rest mass is moving with a velocity of light . Then the wavelength of the de Broglie wave associated with it is :2010Matter Waves

The de-Broglie wavelength of a proton (charge =1.6×10-19 C,m=1.6×10-27 kg) accelerated through a potential difference of 1 kV is

Solution

Relevant Questions

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{- 19} C, m = 1.6 \times 10^{- 27} kg$ ) accelerated through a potential difference of $1 kV$ is