Question

The electron energy in hydrogen atom is given by \(E_n = \frac {(–2.18×10^{–18})}{n^2} \ J\). Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Answer 1

Given

• Energy of electron in hydrogen atom: \[ E_n = \frac{-2.18 \times 10^{-18}}{n^2}\,\text{J} \]
• We need energy to remove the electron completely from \( n = 2 \) to \( n = \infty \).

1. Energy Required from n = 2

Energy at \( n = 2 \):

\[ E_2 = \frac{-2.18 \times 10^{-18}}{2^2} = \frac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19}\,\text{J} \]

Energy at \( n = \infty \): \( E_\infty = 0 \,\text{J} \).

Energy required for ionisation:

\[ \Delta E = E_\infty - E_2 = 0 - (-5.45 \times 10^{-19}) = 5.45 \times 10^{-19}\,\text{J} \]

Energy required = \( 5.45 \times 10^{-19}\,\text{J} \)

2. Longest Wavelength of Light (in cm)

Use the relation: \[ \Delta E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{\Delta E} \] where \( h = 6.626 \times 10^{-34}\,\text{J s} \), \( c = 3.0 \times 10^{8}\,\text{m s}^{-1} \).

\[ \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}} {5.45 \times 10^{-19}} = \frac{1.9878 \times 10^{-25}} {5.45 \times 10^{-19}} \approx 3.65 \times 10^{-7}\,\text{m} \]

Convert metres to centimetres:

\[ \lambda = 3.65 \times 10^{-7}\,\text{m} = 3.65 \times 10^{-7} \times 100\,\text{cm} = 3.65 \times 10^{-5}\,\text{cm} \]

Final Answers

• Energy required to remove the electron from \( n = 2 \): \( 5.45 \times 10^{-19}\,\text{J} \)
• Longest wavelength of light that can cause this transition: \( \lambda \approx 3.6 \times 10^{-5}\,\text{cm} \)