Question

What will be de Broglie wavelength of photon if its accelerated by potential difference of 230 V?

 

 

A proton is accelerated through 230 V. Its de Broglie wavelength is

1. 0.1 nm
2. 0.2 nm
3. 0.3 nm
4. 0.4 nm 

Answer 1

Correct Answer - Option 2 : 0.2 nm

CONCEPT:

Dual nature of matter:

• According to de Broglie, the matter has a dual nature of wave-particle.
• The wave associated with each moving particle is called matter waves.
• ​​de Broglie wavelength associated with the particle

\(⇒ λ = \frac{h}{mv}= \frac{h}{\sqrt{2mE}}\)   

Where, h = Planck's constant, m = mass of a particle, v = velocity of a particle, and E = kinetic energy of the particle

calculation:

Given V = 230 V, q = 1.6 × 10^-19 C, h = 6.6 × 10^-34 J-sec and m = 1.67 × 10^-27 kg

• We know that if a charge of q coulombs is moved through a potential difference of V volts then the kinetic energy gained by the charge is given as,

⇒ E = qV     -----(1)

So the de Broglie wavelength associated with the proton when it is accelerated through 225 V is given as,

\(⇒ λ =\frac{h}{\sqrt{2mE}}\)

\(⇒ λ =\frac{h}{\sqrt{2mqV}}\)

\(⇒ λ =\frac{6.6×10^{-34}}{\sqrt{2×1.67×10^{-27}×1.6×10^{-19}×230}}\)

⇒ λ = 0.19 × 10^-9 m

⇒ λ = 0.19 nm

• Hence, option 2 is correct.