Answer to Question #154038 in Physical Chemistry for Trisha Lalbeharry

Question

 An electron in the ground state of a silicon atom has the quantum numbers (3, 1, -1, +1⁄2) and absorbs infrared radiation of wavelength 1095 nm.

(i) To which energy level does the electron move?

(ii) Calculate the frequency of this radiation.

Answer 1

i)

$v$ = R ( $\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$ ) , $v$ = frequency ,R = 109700 , wavelength =1095nm = 1095×10^-7

$v$ = 1/λ

1/λ = R($\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$ )

$\frac{1}{1095×10^-⁷}$ = 109700 ( $\frac{1}{3²}$ - $\frac{1}{n_2^2}$ )

$\frac{1}{1095×10^-⁷}$ = 109700 ($\frac{n_2^2-9}{9n_2^2}$ )

$\frac{1}{1095×10^-⁷×109700}$ = $\frac{n_2^2-9}{9n_2^2}$

$\frac{1}{12.01215}$ = $\frac{n_2^2-9}{9n_2^2}$

9n$_2^2$ = 12.01215(n$_2^2$ - 9)

9n$_2^2$ = 12.01215n$_2^2$ - 108.10935

12.01215n$_2^2$ - 9n$_2^2$ = 108.10935

3.01215 n$_2^2$ = 108.10935

n$_2^2$ = 108.10935/3.01215

n$_2^2$ = 36

n₂ = 6

ii) $v$ = c/λ

$v$ = (3×10⁸) / (1095×10^-9)

$v$ = ( 3× 10¹⁷ ) / 1095

$v$ = 2.73 × 10¹⁴