Question

When an electron jumps from $n_1^{th}$ orbit to $n_2^{th}$ orbit then the formula for energy radiated out is

• A. $E_1-E_2=hv$
• B. $E_2-E_1=hv$
• C. $hv=\frac{E_1+E_2}{2}$
• D. $hv=\frac{E_1}{E_2}$

Answer 1

The correct option is B $E_2-E_1=hv$

The energy and distance of an electron from nucleus are given in terms of principal quantum number n.
$E_n=-\frac{13.6}{n^2}$ and $r_n=0.529n^2$
Since, $E\propto \frac{1}{n^2}$ and $r_n\propto n^2$, as n increases radius of orbit of revolution of electron i.e. distance of an electron from nucleus increases and magnitude of energy decreases. Hence, for n = 1,2,3,..........so on.
$E_1>E_2>E_3>............$
but there is a negative sign in the formula so when an electron jumps from $n_1^{th}$ orbit to $n_2^{th}$ orbit then the formula for energy radiated out is
$E_2-E_1=h\nu$