first exitation energy of hydrogen : second exitation energy of helium : third exitation of Lithium
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Answer:
12th
Physics
Atoms
Bohr's Model
Energy liberated in the de ...
PHYSICs
Energy liberated in the de-excitation of hydrogen atom from the third level to the first level falls on a photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen-like gas, excited to the second energy level, it is found that the de Broglie wavelength of the fastest photoelectrons now ejected has decreased by factor of 3. For this new gas, difference of energies of the second Lyman line and the first Balmer line is found to be 3 times the ionization potential of the hydrogen atom. Select the correct answers.
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November 22, 2019
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Reshma Prasad
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ANSWER
Second Lyman line: n=3 to n=1
First Balmer line: n=3 to n=2
Difference in their energies is: E
2
−E
1
which is, (13.6−3.4)Z
2
and given that it is equal to 3 times I.E of hydrogen.
⇒ 10.2Z
2
=3×13.6.
Therefore, Z=2
Therefore, the gas is Helium.
Energy liberated by the de-excitation of hydrogen from $$n=3ton=1is,13.6-1.5=12.1\ eV$$
It is now exposed to helium excited to second energy level.
Energy released from helium is E
2
−E
1
=40.8eV
Second time, emitted photo electrons had de-Broglie wavelength decreased by factor 3, which is λ→
3
2λ
⇒K.E→
2
3K.E
Therefore,
12.1=Φ+K.E⟶(i)
40.8=Φ+
2
3K.E
⟶(ii)
Solving these equations, we get Φ=8.5 eV.