Answer to Question #247378 in General Chemistry for Jay

Question

Determine the wavelength in nm of light emitted by an electron as it falls from the 4th to the 2nd energy level.

Answer must be in computer format for scientific notation, e.g. 5.4E5

Answer 1

The expression for the wavelength of radiation is, 

$\frac{1}{\lambda}= R(\frac{1}{n^2_1} ​−\frac{1}{n^2_2} ​)$ λ

Substitute values in the above expression,

$\frac{1}{λ} ​=109677×(\frac{1}{2^2} ​−\frac{1}{4^2} ​)=109677×(\frac{1}{4}-\frac{ 1}{ ​16} ​)$

$=109677×(\frac{12}{64} ​)$

$λ=4.86×10^{−5}$

$cm=486×10^{−9} .$

$m=486nm$