CBSE Class 12 Physics 2014 Delhi Set 2 Paper

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Question : 6
Total: 8
A 12.9eV beam of electronic is used to bombard gaseous hydrogen at room temperature.
Upto which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Paschen series and first member of Balmer series.
Solution:  
Energy of the electron in the nth state of an atom Here, Z is the atomic number of the atom.
For hydrogen atom, Z=1
Energy required to excite an atom from initial state (ni) to final state (nf),
E=13.6(
1
nf2
1
ni2
)
eV

13.6
nf2
+
13.6
ni2
=12.9

This energy must be equal to or less than the energy of the incident electron beam.
13.612.9=
13.6
nf2
[n=1]

nf=4.4
State cannot be a fraction number.
nf=4
Hence, the hydrogen atom would be excited up to 4th energy level.
Rydberg's formula for the spectrum of the hydrogen atom is given by:
1
λ
=R(
1
n12
1
n22
)

Here, λ is the wavelength
Rydberg's canstant, R=1.097×107m1
For the first member of the Paschen series
1
λ
=1.097×107(
1
32
1
42
)

λ=18752.4
For the first member of Balmer series
n1=2,n2=3
1
λ
=1.097×107(
1
22
1
32
)

λ=6563.3
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