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 = λ = ?
Solution:
The formula for wavelength is
λ = `\frac{h}{sqrt {2mE_n}}`
Where
λ = `\frac{h}{sqrt {frac{2m2Ï€²k²me⁴}{n²h²}}}`
λ = `\frac{h}{frac{2mÏ€ke²}{nh}}`
or
λ = `\frac{nh²}{2mÏ€ke²}`
by putting the corresponding values
λ = `\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}`
λ = `\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]
λ = `\frac{175.2976x 10⁻⁶⁸ m⁴ kg² s⁻²}{1318.80849408 x 10⁻⁶⁰ kg m s⁻² x kg m² }`
or
λ = 0.1329x 10⁻⁸ m
or
λ = 1.329x 10⁻⁹ m
Or
λ = 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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