Question 109: Dual Nature of Matter and Radiation

Question

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :
(Assume $$m_{p}=m_{e} 	imes 1849$$ )

Accepted Answer

The de Broglie wavelength (ÃÂ») 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:
$$
m_p = 1849 	imes m_e
$$
Let's find the ratio of the de Broglie wavelengths of the proton ($ÃÂ»_p$) and the electron ($ÃÂ»_e$):
$$
\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}}
$$
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:
$$
\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 	imes m_e}}
$$
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.

Answer

1:43

Answer

1:62

Answer

2:43

Answer

1:30

Dual Nature of Matter and Radiation

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy : (Assume $m_{p}=m_{e} \times 1849$ )

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