Question

A proton and an α particle are accelerated through the same potential difference. Which one of the two has (i) greater de- Broglie wavelength, and(ii) less kinetic energy? Justify your answer.

Answer 1

Proton has a greater de-Broglie wavelength than the alpha particle and proton has less value of Kinetic energy.

Explanation:

Using de-Broglie wavelength formula, we already know that proton and alpha particle are accelerated through the same potential.

So, both their velocities are same.

λ=  h/√ 2meV

λ ∝ 1 / ​√mq

λα / λp =   √ mpqp / maqa

Charge on alpha particle = 2 x charge on proton

Mass of alpha particle = 4 x mass of proton

λα / λp = 1/2 √2

λp  = 2√2λα

Proton has a greater de-Broglie wavelength than the alpha particle.

(b) Kinetic energy is directly proportional to the q.

As the charge of alpha particle is more than that of proton

Hence proton has less value of Kinetic energy.

Thus Proton has a greater de-Broglie wavelength than the alpha particle and proton has less value of Kinetic energy.

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Answer 2

(i) Greater de-Broglie wavelength:

From question, we know that the proton and an α particle are accelerated through the same potential difference. So, the velocities of the particles remains same.

On using de-Broglie wavelength formula, we get,

$\lambda=\frac{h}{\sqrt{2 m V_{0} q}}$

Now, on taking the constants out, we get,

$\lambda \alpha \frac{1}{\sqrt{\mathrm{mq}}}$

On diving the wavelengths of proton and an α particle, we get,

$\frac{\lambda_{\alpha}}{\lambda_{p}}=\sqrt{\frac{m_{p} q_{p}}{m_{\alpha} q_{\alpha}}}$

We know that,

Charge on alpha particle = 2 times charge on proton

Mass of alpha particle = 4 times mass of proton

Now, on substituting the values in above equation, we get,

$\frac{\lambda_{\alpha}}{\lambda_{p}}=\frac{1}{2 \sqrt{2}}$

$\lambda_{p}=2 \sqrt{2} \lambda_{\alpha}$

(ii) Less kinetic energy:

When the acceleration is in same potential then, the kinetic energy of the particle is directly proportional to the charge of the particle.

Since, the charge of the alpha particle is more than proton, then the kinetic energy of the alpha particle is higher compared to proton.

$K.E \ of \alpha \ particle > K.E \ of \ proton$