Question

Proton and electron have equal kinetic energy, the ratio of de-Broglie wavelength of proton and electron is $\frac{1}{x}$. Find x. (Mass of proton 1849 times mass of electron)

Answer 1

The de Broglie wavelength of a particle is given by:
$λ = \frac{h}{ p}$
where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.
The momentum of a particle is given by:
p = mv
where m is the mass of the particle and v is its velocity.
Let K be the kinetic energy of both proton and electron. Then, we can write:
$\frac{mv^2}{ 2}$ = K
The velocity of the proton and the electron will be different since they have different masses and the same kinetic energy. Let v_p and v_e be the velocities of proton and electron respectively.
For proton, we have:
$v_p = √(\frac{2K}{m_p})$
For electron, we have:
$v_e = √(\frac{2K}{m_e})$
where m_p and m_e are the masses of proton and electron respectively.
The momentum of the proton is given by:
$p_p = m_p v_p$
The momentum of the electron is given by:
$p_e = m_e v_e$
The ratio of the de Broglie wavelengths of proton and electron is given by:
$\frac{λ_p}{ λ_e} = (\frac{p_e }{ p_p}) = \frac{(m_e v_e) }{(m_p v_p)}$
Substituting the expressions for v_p and v_e, we get:
$\frac{λ_p }{ λ_e} = √(\frac{m_e }{ m_p})$
Given that the mass of proton is 1849 times the mass of electron, we have:
$\frac{λ_p}{ λ_e} = √(\frac{m_e}{ (1849 m_e)}) = \frac{1}{43}$
Therefore, the value of x is 43. The ratio of de Broglie wavelengths of proton and electron is $\frac{1}{x} = \frac{1}{43}$.
Answer. 43