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

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$.

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ =  energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$.