Question

emissions are observed at wavelengths of 383.65 and 379.90nm for transitions from excited states of the hydrogen atom to the n=2 state. determine the quantum numbers nh for these emissions

Answer 1

Wavelengths of the two emissions are:

1 =  383.65 nm = 383.65x10^-9 m

2 =  379.90 nm = 379.90x10^-9 m

For the first emission, using rydberg's formula:

1/ = R {(1/n1)² - (1/n2)²}

1 / (383.65x10^-9) = 1.097x10⁷ * {(1/2)² - (1/n2)²}

10⁹ / 383.65 = 1.097x10⁷ * { (1/4) - (1/n2)²}

2.6x10⁶ = 1.097x10⁷ * { (1/4) - (1/n2)²}

2.6 / 10.97 = 0.25 - (1/n2)²  

0.237 = 0.25 - (1/n2)²  

(1/n2)² = 0.013

1/n2 = 0.114

n2 = 9

For the second emission, using rydberg's formula:

1/ = R {(1/n1)² - (1/n2)²}

1 / (379.90x10^-9) = 1.097x10⁷ * {(1/2)² - (1/n2)²}

2.63x10⁶ = 1.097x10⁷ * { (1/4) - (1/n2)²}

2.63 / 10.97 = 0.25 - (1/n2)²  

0.24 = 0.25 - (1/n2)²  

(1/n2)² = 0.01

1/n2 = 0.1

n2 = 10

Therefore, the quantum numbers for these emissions are 9 and 10 respectively.