QJEEM23Ph2c86: JEE Main 2023 Physics - Class 12 JEEM23Ph2c86 Solution: The ratio of the de-Broglie wavelengths of proton and electron…

The de Broglie wavelength (λ) of a particle can be found using the formula:

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

where h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:

$p = \sqrt{2mK}$

Combining these two equations, we get:

$\lambda = \frac{h}{\sqrt{2mK}}$

Now, we are given that the kinetic energy of the proton and electron is the same. Let's denote the masses of the proton and electron as $m_p$ and $m_e$ , respectively. We are given the relationship between the two masses:

$m_p = 1849 \times m_e$

Let's find the ratio of the de Broglie wavelengths of the proton ( $λ_p$ ) and the electron ( $λ_e$ ):

$\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}}$

Simplifying the expression, we get:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m_eK}}{\sqrt{2m_pK}}$ The 2K terms cancel out:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}}$

Substitute the given relationship between the masses:

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 \times m_e}}$

Further simplification:

$\frac{\lambda_p}{\lambda_e} = \frac{1}{\sqrt{1849}}$

Since 1849 is equal to $43^2$ :

$\frac{\lambda_p}{\lambda_e} = \frac{1}{43}$

Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.