In a hydrogen atom, an electron undergoes transition from an energy level with n=4 to an energy
level with n=2. Calculate the wavelength of the spectral line thus obtained. In which region of
electromagnetic spectrum this line would be observed and what would be its colour?
Answers
Answer:
ANSWER
The expression for the wavelength of radiation is,
λ
1
=R(
n
1
2
1
−
n
2
2
1
)
Substitute values in the above expression,
λ
1
=109677×(
2
2
1
−
4
2
1
)=109677×(
4
1
−
16
1
)=109677×(
64
12
)
λ=4.86×10
−5
cm=486×10
−9
m=486nm.
The line belongs to bluish green colour.
Answer:
Wavelength [lamda] = 486nm
The line is observed in { infrared spectrum }
In the colour of blue
Explanation:
Here I used normal brackets ( ) instead of symbol of multiplication . So don't get confused :)
The expression for the wavelength of radiation is,
1/lamda = R ( 1/n1^2 - 1/n2^2)
1/ lamda = 1.097 (10^7) ( 1/2^2 - 1/4^2)
1/ lamda = 1.097 (10^7) {( 4 - 1 ) / 16}
1/ lamda = 1.097 { 10^7 { 3} / 16 }
lamda = 16/ 3.291 ( 10^7 ) m^-1
hence wave length = lamda = 4.816 ( 10^-7 ) m^-1 which is equal to 486nm
It is observed in the region of "Infrared spectrum" In "Blue" colour