Question

Two close lying bands in a UV spectrum occur at 274 nm and 269 nm. The magnitude of the energy gap between the two bands is ____cm^-1. (round off to the nearest integer)

Answer 1

Correct Answer: 670

The energy of light is related to its wavelength by the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant (\(6.626 \times 10^{-34}\) Js), \(c\) is the speed of light (\(3 \times 10^{8}\) m/s), and \(\lambda\) is the wavelength in meters. Given \(\lambda_1 = 274\) nm and \(\lambda_2 = 269\) nm.

To find the energy difference between the two bands, we first convert the wavelengths from nanometers to meters: \(\lambda_1 = 274 \times 10^{-9}\) m and \(\lambda_2 = 269 \times 10^{-9}\) m.

The energy difference \(\Delta E\) in joules is given by:

\(\Delta E = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1}\)

Calculating \(\Delta E\):

\(\Delta E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{269 \times 10^{-9}} - \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{274 \times 10^{-9}}\)

Solving, \(\Delta E \approx 7.396 \times 10^{-19} - 7.263 \times 10^{-19} = 1.33 \times 10^{-20}\) J

Converting this energy difference to wavenumbers (cm^-1):

\(1 \text{ J} = 1 \text{ N m} = 1 \text{ kg m}^2/\text{s}^2\) and \(1 \text{ cm}^{-1} = 100 h c / (\lambda \times 100 \text{ m/s})\)

\(\Delta \nu = \frac{1.33 \times 10^{-20}}{h \times c} \approx \frac{1.33 \times 10^{-20}}{6.626 \times 10^{-34} \times 3 \times 10^{10}}\) cm^-1

Calculating \(\Delta \nu\), we get:

\(\Delta \nu \approx 669.68\) cm^-1

Rounding this to the nearest integer gives \(670\) cm^-1. Thus, the magnitude of the energy gap is 670 cm^-1.