Question

If an electron in \(n=4\) orbit of hydrogen atom jumps to \(n=3\), the energy released and wavelength emitted are:

• A. \(0.66\,eV , 1.88 \times 10^{-6}\,m\)
• B. \(1.89\,eV , 1.98 \times 10^{-7}\,m\)
• C. \(0.29\,eV , 1.78 \times 10^{-5}\,m\)
• D. \(0.98\,eV , 0.93 \times 10^{-6}\,m\)

Hint:

For hydrogen transitions: \[ \lambda(\text{nm}) = \frac{1240}{E(\text{eV})} \] is the fastest way to calculate wavelength.

Answer 1

The Correct Option is A

Concept: Energy of electron in hydrogen atom: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

Step 1:Calculate energies. \[ E_4 = -\frac{13.6}{16} = -0.85\,\text{eV} \] \[ E_3 = -\frac{13.6}{9} \approx -1.51\,\text{eV} \]

Step 2:Energy released. \[ \Delta E = E_3 - E_4 = (-1.51) - (-0.85) = -0.66\,\text{eV} \] \[ |\Delta E| = 0.66\,\text{eV} \]

Step 3:Wavelength calculation. \[ \lambda = \frac{hc}{E} \] Using: \[ \lambda(\text{nm}) = \frac{1240}{E(\text{eV})} \] \[ \lambda = \frac{1240}{0.66} \approx 1879\,\text{nm} = 1.88 \times 10^{-6}\,\text{m} \]