Both an electron and a photon have same de-Broglie wavelength of 1 . 2. `Å` . The ratio of their energies is nearly

Options

A. `1 : 100`
B. `1 : 10`
C. `1 : 1000`
D. `1 : 1`

Answer

A

Explanation

The de-Broglie wavelength of electron
`lambda_(e)= (h)/(sqrt(mK_(e)))`
`K_(e)` = Kenetic energy of electron
`lambda_(e)^(2) = (h^(2))/(lambda^(2)xx 2 m_(e))`
De-Brodlie wavelength of photon is `lambda_(p)` Kinetic energy of photon,
`K_(p) = hv = (hc)/(lambda_(p))`
`:. (K_(e))/(K_(p)) = (h^(2) //lambda_(e)^(2) . 2m_(e))/(hc//lambda_(p))`
`lambda_(e) = lambda_(p)`
`(K_(e))/(K_(p)) = (lambda_(p))/(hc) xx (h^(2))/(lambda_(e)^(2) xx 2m_(e))`
We know that `m_(e) = 9.1 xx 10^(-31) kg`
` h = 6 . 66 xx 10^(-34) J - sec`
`lambda_(e) = 12 Å`
`= 12 xx 10^(-10) m`
and `c = 3 xx 10^(8) m//s`
Then, `(K_(e))/(K_(p)) = (h)/(c lambda_(e) xx 2 m_(e))`
` = (6 . 66 xx 10^(-34))/(3 xx 10^(8) xx 12 xx 10^(-10) xx 2 xx 9 . 1 xx 10^(-31))`
`~=(1)/(100)`
`:. K_(e) :K_(p) : : 1 : 100`