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Question : 11
Total: 11
(a) (i) Write the limitations of Rutherford's model of atom.
(ii) The wavelength of the second line of the Balmer series in the hydrogen spectrum is4 8 6 1 Å
OR
(b) (i) Increase in the intensity of the radiation causing photo-electric emission from a surface, does not affect the maximum K.E. of the photo electrons. Explain.
(ii) The photon emitted during the de-excitation from the first excited level to the ground state of hydrogen atom is used to irradiate a photo cathode in which stopping potential is5 V
(ii) The wavelength of the second line of the Balmer series in the hydrogen spectrum is
OR
(b) (i) Increase in the intensity of the radiation causing photo-electric emission from a surface, does not affect the maximum K.E. of the photo electrons. Explain.
(ii) The photon emitted during the de-excitation from the first excited level to the ground state of hydrogen atom is used to irradiate a photo cathode in which stopping potential is
Solution: 👈: Video Solution
(a) (i) Limitations of Rutherford's model of atom:
● Rutherford's model failed to explain the stability of atoms.
Charged particle that is moving in a circular path undergoes an acceleration and radiates energy. Thus, the revolving electron losses energy and finally falls into the nucleus. But this is not the reality. Atoms are stable enough.
● Atom exhibits line spectrum. Rutherford's model failed to explain this. According to Rutherford, if electron continuously emits energy then there should be continuous band spectrum. But that is not the reality.
(ii) For Balmer series
= R (
−
)
For2 nd n = 4
∴
= R (
−
)
Or,
=
∴ R =
For1 st n = 3
∴
= R (
−
)
Or,
=
Or, 1 ∕ λ ′ =
Or, λ ′ =
∴ λ ′ = 6562.35 Å
OR
(b) (i) From Einstein's photoelectric equation.
h v = ϕ 0 + K E max
So, KE energy of photoelectron depends on the frequency of incident radiation and the work function of the substance. There is no intensity term in the equation. Hence, the KE is independent of the intensity of the radiation.
(ii) From Einstein's photoelectric equation,
e V s = h v − φ 0
andh v = E 2 − E 1
= −
− ( −
)
= 10.2 eV
Putting the Einstein's equation,
5 eV = 10.2 eV − ϕ 0
∴ ϕ 0 = 5.2 eV
● Rutherford's model failed to explain the stability of atoms.
Charged particle that is moving in a circular path undergoes an acceleration and radiates energy. Thus, the revolving electron losses energy and finally falls into the nucleus. But this is not the reality. Atoms are stable enough.
● Atom exhibits line spectrum. Rutherford's model failed to explain this. According to Rutherford, if electron continuously emits energy then there should be continuous band spectrum. But that is not the reality.
(ii) For Balmer series
For
For
OR
(b) (i) From Einstein's photoelectric equation.
So, KE energy of photoelectron depends on the frequency of incident radiation and the work function of the substance. There is no intensity term in the equation. Hence, the KE is independent of the intensity of the radiation.
(ii) From Einstein's photoelectric equation,
and
Putting the Einstein's equation,
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