Question

The electron energy in hydrogen atom is given by $E = - \dfrac{{21.7 \times {{10}^{ - 12}}}}{{{n^2}}}erg$ . Calculate the energy required to remove an electron completely from $n = 2$ orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition?

Answer 1

Hint: In order to answer this question, first we will rewrite the given number of orbit and the electron energy in the hydrogen atom and then we will apply the formula of electron energy in the terms of wavelength and the speed of light, i.e.. $E = \dfrac{{hc}}{\lambda }$ .

Complete answer:
Energy required to remove electrons completely from $n = 2$ orbit.
And, the electron energy in hydrogen atom, for $n = 2$ :
$
  E = - \dfrac{{21.7 \times {{10}^{ - 12}}}}{{{2^2}}}erg \\
  \,\,\,\,\, = - 5.425 \times {10^{ - 12}}erg \\
 $
As we know that, ${E_\infty } = 0$
So, $\Delta E = change\,in\,energy = {E_\infty } - E = 0 - ( - 5.425 \times {10^{ - 12}})erg = 5.425 \times {10^{ - 12}}erg$
Now, we will apply the formula of energy in terms of wavelength and the speed of light:
$\because \Delta E = \dfrac{{hc}}{\lambda } = 5.425 \times {10^{ - 12}}$
where, $h$ is the planck's constant, whose value is $6.626 \times {10^{ - 34}}J.H{z^{ - 1}}$ .
$c$ is the speed of light, whose value is $3 \times {10^8}m.{s^{ - 1}}$ .
$
   \Rightarrow \dfrac{{{{10}^7} \times 6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times {{10}^2}}}{\lambda } = 5.425 \times {10^{ - 12}} \\
   \Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}}}}{{5.425 \times {{10}^{ - 12}}}} \\
  \,\,\,\,\,\,\,\,\,\,\, = 3.664 \times {10^{ - 5}}cm = 366.40nm \\
 $
Hence, the longest wavelength (in cm) of light that be used to cause this transition $3.664 \times {10^{ - 5}}cm$ .

Note:
The wavelength of visible light is \[400nm{\text{ }}to{\text{ }}700nm\] , and this is where we learn about the wavelengths of different colours in the visible spectrum of light. The visible light spectrum has a large number of various colours with different wavelengths.