Find the wavelength associated with an electron in the state n=4 of hydrogen. (Answer: 1.33 nm) 



Given:

Atom: Hydrogen

Energy level n = 4

∴ Plank's Constant = h =  6.62 x 10⁻³⁴ m² kg s⁻¹

∴ Mass of the electron = m = 9.11 x 10⁻³¹ kg

∴ Charge on the electron = e = 1.6 x 10⁻¹⁹ C

∴ Coulomb's Constant  = K = 9 x 10⁹ N m² C⁻²


To Find:


The wavelength associated with electron Î» =  ?


Solution:

The  formula for wavelength is 

 Î» = `\frac{h}{sqrt {2mE_n}}`

Where

`\E_n` = 
`\frac{2Ï€²k²me⁴}{n²h²}`  So,

 Î» = `\frac{h}{sqrt {frac{2m2Ï€²k²me⁴}{n²h²}}}`

 Î» = `\frac{h}{frac{2mÏ€ke²}{nh}}`

or

 Î» = `\frac{nh²}{2mÏ€ke²}`

by putting the corresponding values 



 Î» = `\frac{4 x (6.62 x 10⁻³⁴ m² kg s⁻¹)²}{2x9.11 x 10⁻³¹ kg x 3.1416 x 9 x 10⁹ N m² C⁻² x(1.6 x 10⁻¹⁹ C)²}`


 Î» `\frac{4 x 43.8244 x 10⁻⁶⁸ m⁴ kg² s⁻²}{515.159568 x 10⁻²² kg N m² C⁻² x2.56 x 10⁻³⁸ C²}`

[N = kg m s⁻²]

 Î» = `\frac{175.2976x 10⁻⁶⁸ m⁴ kg² s⁻²}{1318.80849408 x 10⁻⁶⁰ kg m s⁻² x kg m² }`

or

 Î» = 0.1329x 10⁻⁸ m

or

 Î» = 1.329x 10⁻⁹ m

Or

 Î» = 1.329 nm  ---------------Ans.

Thus the wavelength associated with an electron in the state n=4 of hydrogen is 121.9  nm  



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