Question

Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonrelativistic.

Answer 1

Answer:

the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

Explanation:

We know, wavelength is expressed in terms of Kinetic Energy by :

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Therefore , $E=\dfrac{h^2}{2 \lambda^2 m}$

It is given that both electron and proton have same wavelength.

Therefore,

$E_e=\dfrac{h^2}{2 \lambda^2 m_e}$   .... equation 1.

$E_p=\dfrac{h^2}{2 \lambda^2 m_p}$   .... equation 2.

Now, dividing equation 1 by 2 .

We get ,

$\dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}$

Putting value of mass of electron = $9.1\times 10^{-31}\ kg$ and mass of proton = $1.67\times 10^{-27}\ kg.$

We get :

$\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16$

Hence , this is the required solution.

Answer 2

Answer:

$\frac{KE_e}{KE_p}=1835.16$

Explanation:

Given that the wavelengths of electron and proton are equal at non- relativistic speed.

From De-Broglie wave equation we know that:

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

where:

$\lambda=$ wavelength

$h=$ Planck's constant

$p=$ linear momentum of  the particle

Then'

$\lambda_e=\lambda_p$

$\frac{h}{p_e} =\frac{h}{p_p}$

$\frac{1}{m_e.v_e} =\frac{1}{m_p.v_p}$ ..................................(1)

we've mass of electron, $m_e=9.1\times 10^{-31}\ kg$

mass pf proton, $m_p=1.67\times 10^{-27}\ kg$

Now,

kinetic energy of electron:

$KE_e=\frac{1}{2} m_e.v_e^2$

kinetic energy of proton:

$KE_p=\frac{1}{2}m_p.v_p^2$

So,

$\frac{KE_e}{KE_p}=\frac{m_e.v_e^2}{m_p.v_p^2}$

from eq. (1)

$\frac{KE_e}{KE_p}=\frac{m_e}{m_p} \times \frac{m_p^2}{m_e^2}$

$\frac{KE_e}{KE_p}= \frac{m_p}{m_e}$

$\frac{KE_e}{KE_p}=\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}$

$\frac{KE_e}{KE_p}=1835.16$