Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n = 2 and the atom absorbs a photon of light with a frequency of 4.57 ×10 14 Hz.

Answer 1

From Rydberg's Equation , 1 /w = R [ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ]
Where R = Rydberg's constant = 10.96 * 10^6 m^-1
w = wavelength = ?
n = ?
n' = 4
w = wave length = 486 nm = 486*10^-9 m
putting the values we get
1 /w = R [ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ]
[ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ] = 1/RW
[ ( 1/ n^ 2 ) - ( 1/ 4 ^2 ) ] = 1/(10.96*10^6*486*10^-9)
[ ( 1/ n^ 2 ) - 0.0625 = 0.1877
1/ n^ 2 = 0.25
n = 2
final value of n in a hydrogen atom transition is 2..

Answer 2

From Rydburg's Equation , 1 /w = R [ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ]
Where R = Rydburg's constant = 10.96 * 10^6 m^-1
w = wavelength = ?
n = ?
n' = 4
w = wave length = 486 nm = 486*10^-9 m
Plug the values we get
1 /w = R [ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ]
[ ( 1/ n^ 2 ) - ( 1/ n' ^2 ) ] = 1/RW
[ ( 1/ n^ 2 ) - ( 1/ 4 ^2 ) ] = 1/(10.96*10^6*486*10^-9)
[ ( 1/ n^ 2 ) - 0.0625 = 0.1877
1/ n^ 2 = 0.25
---> n^2 = 1/0.25 = 4
----> n = 2
So end final value of n in a hydrogen atom transition is 2