Question

Calculate the energy required to excite the hydrogen electron from n = 1 to level n = 2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state. What kind of electromagnetic radiation is used?

Answer 1

Answer:

The energy required is 1.634×$10^{-18}$J, the wavelength is 1.215×$10^{-7}$m, and the kind of electromagnetic radiation is the ultraviolet radiation.

Explanation:

In the question above, the initial level $n_{initial}$ = 1, and $n_{final}$ = 2. Then, we use Rydberg's equation to calculate the wavelength of the   light absorbed by the atom during the transition.

1/wavelength = R($\frac{1}{n_{initial} ^{2} } -\frac{1}{n_{final} ^{2} }$)

where:

R is the Rydberg's constant which is equal to 1.0974×$10^{7}$ $m^{-1}$

Therefore,

1/wavelength = 1.0974×$10^{7}$ $m^{-1}$×($\frac{1}{1 ^{2} } -\frac{1}{2^{2} }$) = 1.0974×$10^{7}$ $m^{-1}$×$(\frac{1}{4}-\frac{1}{1} )$ = (-) 0.8228*10^{7}

Thus,

wavelength = $\frac{1}{0.8228*10^{7}}$ = 1.215*$10^{-7} \frac{1}{m}$

The energy required (E) can be calculated using:

E = h*c/wavelength

where:

h is the Planck's constant = 6.626*$10^{-34} J.s$

c is the speed of light = 299,792,458 m/s

Therefore,

E = 6.626*$10^{-34} J.s$*299,792,458 m/s÷ 1.215*$10^{-7} \frac{1}{m}$ = 1.634*$10^{-18} J$

The kind of electromagnetic radiation used is  is the ultraviolet radiation. This is the type of electromagnetic radiation with wavelength between 10 nm and 400 nm