The de-broglie wavelength of a photon is twice the de-broglie wavelength of an electron. The speed of the electron is `v_(e)=c/100`. Then

Options

A. `E_e/E_p=10^(-4)`
B. `E_e/E_p=10^(-2)`
C. `p_e/(m_ec)=10^(-2)`
D. `p_e/(m_ec)=10^(-4)`

Answer

B, C

Explanation

Given, velocity of electron , `v_e = c//100`
The de Broglie wavelength of electron is given by
`lamda_e=h/(m_ev_e)=h/(m_e(c//100))=(100h)/(m_ec)" "...(i)`
Kinetic energy of electron will be
`E_e=(1/2)m_ev_e^2`
`implies m_ev_e=sqrt(2E_em_e)`
`:. lamda_e=h/(m_ev_e)=h/(sqrt(2m_eE_e))`
`implies E_e=h^2/(2lamda_e^2m_e)`
Let wavelength of photon be `lamda_p` . Given,
`lamda_p=2lamda_e`
Then energy of photon is
`E_p=(hc)/lamda_p=(hc)/(2lamda_e)`
`impliesE_p/E_e=(hc)/(2lamda_e)xx(2lamda_e^2m_e)/h^2=(lamda_em_ec)/h`
Using (i) , we have
`E_p/E_e=(100h)/(m_ec)xx(m_ec)/h=100`
`impliesE_p/E_e=1/100=10^(-2)`
Momentym of electron is given by
`p_e=m_ev_e=m_exx(c//100)`
`impliesp_e/(m_ec)=1/100=10^(-2)`