Question

In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted? electron falls from the fourth to the second orbital, what from the A. 3.04 x 10 Hz (or cycles per second) B. 608 x 1014 ã C. 3.04 x 1015 Hz D. 4.00 x 10'"Hz E. 4.00 x 10 Hz

Answer 1

The frequency of the photon is
`3.069*10^(14)\ Hz`.

Step-by-step explanation:

Given that,

Energy
`E=4.07*10^(-19)\ J`

We need to calculate the energy

Using relation of energy

`E_(4)-E_(2)=\Delta E`

Where,
`\Delta E` = energy spacing

`4h\\u-2h\\u=4.07*10^(-19)`

`\\u=(4.07*10^(-19))/(2h)`

Put the value of h into the formula

`\\u=(4.07*10^(-19))/(2*6.63*10^(-34))`

`\\u=3.069*10^(14)\ Hz`

Hence, The frequency of the photon is
`3.069*10^(14)\ Hz`.