QJEEM24Ph97d0: JEE Main 2024 Physics - Class 12 JEEM24Ph97d0 Solution: The de Broglie wavelengths of a proton and an particle are and…

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

$\lambda = \frac{h}{p}$

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$ :

$p = mv$

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

$\lambda = \frac{h}{mv}$

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$ :

$\lambda_p = \frac{h}{m_p v_p}$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$ :

$2\lambda = \frac{h}{m_\alpha v_\alpha}$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_\alpha}$ . Using the given data about wavelengths:

$\lambda_p = \lambda$

$\lambda_\alpha = 2\lambda$

Using the de Broglie equation for both particles:

$\frac{h}{m_p v_p} = \lambda$

$\frac{h}{m_\alpha v_\alpha} = 2\lambda$

Dividing the second equation by the first equation gives us:

$\frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda}$

$\frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1}$

$\frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p}$

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ( $m_\alpha \approx 4m_p$ ).

Substituting $m_\alpha$ with $4m_p$ in the equation:

$\frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p}$

$\frac{v_p}{v_\alpha} = 2 \cdot 4$

$\frac{v_p}{v_\alpha} = 8$

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$ , which corresponds to Option A.