Respuesta :
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 :
[tex]\lambda=\dfrac{h}{\sqrt{2mE}}[/tex]
Therefore , [tex]E=\dfrac{h^2}{2 \lambda^2 m}[/tex]
It is given that both electron and proton have same wavelength.
Therefore,
[tex]E_e=\dfrac{h^2}{2 \lambda^2 m_e}[/tex] .... equation 1.
[tex]E_p=\dfrac{h^2}{2 \lambda^2 m_p}[/tex] .... equation 2.
Now, dividing equation 1 by 2 .
We get ,
[tex]\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}[/tex]
Putting value of mass of electron = [tex]9.1\times 10^{-31}\ kg[/tex] and mass of proton = [tex]1.67\times 10^{-27}\ kg.[/tex]
We get :
[tex]\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16[/tex]
Hence , this is the required solution.
Answer:
[tex]\frac{KE_e}{KE_p}=1835.16[/tex]
Explanation:
Given that the wavelengths of electron and proton are equal at non- relativistic speed.
From De-Broglie wave equation we know that:
[tex]\lambda =\frac{h}{p}[/tex]
where:
[tex]\lambda=[/tex] wavelength
[tex]h=[/tex] Planck's constant
[tex]p=[/tex] linear momentum of the particle
Then'
[tex]\lambda_e=\lambda_p[/tex]
[tex]\frac{h}{p_e} =\frac{h}{p_p}[/tex]
[tex]\frac{1}{m_e.v_e} =\frac{1}{m_p.v_p}[/tex] ..................................(1)
we've mass of electron, [tex]m_e=9.1\times 10^{-31}\ kg[/tex]
mass pf proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Now,
kinetic energy of electron:
[tex]KE_e=\frac{1}{2} m_e.v_e^2[/tex]
kinetic energy of proton:
[tex]KE_p=\frac{1}{2}m_p.v_p^2[/tex]
So,
[tex]\frac{KE_e}{KE_p}=\frac{m_e.v_e^2}{m_p.v_p^2}[/tex]
from eq. (1)
[tex]\frac{KE_e}{KE_p}=\frac{m_e}{m_p} \times \frac{m_p^2}{m_e^2}[/tex]
[tex]\frac{KE_e}{KE_p}= \frac{m_p}{m_e}[/tex]
[tex]\frac{KE_e}{KE_p}=\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}[/tex]
[tex]\frac{KE_e}{KE_p}=1835.16[/tex]
