Question

What will be the ratio of de-Broglie wavelength (λα:λp) when an alpha-particle and a proton are accelerated from rest by a potential difference of V volts and 8 V volts respectively?

Answer 1

1:1

Answer 2

To determine the ratio of de-Broglie wavelengths, we first recall the relevant formulas:

1. De-Broglie Wavelength: $\lambda = \frac{h}{p}$ where $h$ is Planck's constant and $p$ is the momentum of the particle.

2. Kinetic Energy from Potential Difference: When a charged particle is accelerated from rest by a potential difference $V_{accel}$, its kinetic energy (KE) is given by: $KE = qV_{accel}$ where $q$ is the charge of the particle.

3. Momentum and Kinetic Energy Relation: The kinetic energy is also related to momentum and mass ($m$) by: $KE = \frac{p^2}{2m}$ From this, we can express momentum as: $p = \sqrt{2mKE}$

Combining these equations, we can express the de-Broglie wavelength in terms of mass, charge, and accelerating potential difference: $p = \sqrt{2m(qV_{accel})}$ Substituting this into the de-Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mqV_{accel}}}$

Now, let's apply this formula to the alpha-particle ($\alpha$) and the proton (p):

For the alpha-particle:

• Mass of alpha-particle, $m_\alpha = 4m_p$ (where $m_p$ is the mass of a proton)
• Charge of alpha-particle, $q_\alpha = 2e$ (where $e$ is the charge of a proton)
• Accelerating potential difference for alpha-particle, $V_\alpha = V$

The de-Broglie wavelength for the alpha-particle is: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}}$ $\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)(V)}}$ $\lambda_\alpha = \frac{h}{\sqrt{16m_peV}}$

For the proton:

• Mass of proton, $m_p = m_p$
• Charge of proton, $q_p = e$
• Accelerating potential difference for proton, $V_p = 8V$

The de-Broglie wavelength for the proton is: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}}$ $\lambda_p = \frac{h}{\sqrt{2(m_p)(e)(8V)}}$ $\lambda_p = \frac{h}{\sqrt{16m_peV}}$

Ratio of de-Broglie wavelengths ($\lambda_\alpha : \lambda_p$): Now, we find the ratio of their de-Broglie wavelengths: $\frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{16m_peV}}}{\frac{h}{\sqrt{16m_peV}}}$ $\frac{\lambda_\alpha}{\lambda_p} = 1$

Therefore, the ratio of the de-Broglie wavelengths is $1:1$.