Question

An electron in hydrogen atom stays in 2nd orbit for ${10}^{-8}$ . How many, revolutions will it make till it jumps to the ground state ?

Answer 1

As the $e^{⊝}$ has to jump from $2^{nd}$ orbit to $1^{st}$ orbit

$v=\frac{\Delta E}{h}=\frac{E_2-E_1}{h}$

$v=\frac{(-3.4)-(-13.6)}{6.63\times {10}^{-34}}$

$v=\frac{10.2\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}$

$v=2.46\times {10}^{15}$Hz

$\Rightarrow$ It means that in $1$ sec $e^{⊝}$ completes $2.46\times {10}^{15}$ recolutions.

Then, in ${10}^{-8}s$ the electron will undergo

$2.46\times {10}^{15}\times {10}^{-8}$ revolutions

$2.46\times {10}^7$ revolutions.