Question

What will be the ratio of proton and alpha particle if both are accelerated using the same potential difference?

1. 2 : 1
2. 1 : 1
3. 1 : 2
4. 2√2 : 1

Answer 1

Correct Answer - Option 4 : 2√2 : 1

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:

Let V = potential difference

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

⇒ E = qV     -----(1)

So the energy gained by the proton due to potential difference is given as (q_P = e = 1.6 × 10^-19 C),

⇒ E_P = eV     -----(2)

And the energy gained by the α particle due to potential difference is given as (qα = 2e = 3.2 × 10-19 C),

⇒ Eα = 2eV     -----(3)

The de Broglie wavelength of the proton is given as (m_P = m, E_P = eV),

\(\Rightarrow \lambda_P=\frac{h}{\sqrt{2m_{P}E_P}}\)

\(\Rightarrow \lambda_P=\frac{h}{\sqrt{2meV}}\)     -----(4)

The de Broglie wavelength of the α particle is given as (mα = 4m, Eα = 2eV),

\(\Rightarrow \lambda_\alpha=\frac{h}{\sqrt{2m_{_\alpha}E_\alpha}}\)

\(\Rightarrow \lambda_\alpha=\frac{h}{\sqrt{2\times 4m\times2eV}}\)

\(\Rightarrow \lambda_\alpha=\frac{h}{\sqrt{8\times2meV}}\)     -----(5)

By equation 4 and equation 5,

\(\Rightarrow \frac{\lambda_P}{\lambda_\alpha}=\frac{h}{\sqrt{2meV}}\times\frac{\sqrt{8\times2meV}}{h}\)

\(\Rightarrow \frac{\lambda_P}{\lambda_\alpha}=\frac{2\sqrt{2}}{1}\)

• Hence, option 4 is correct.