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Grade: upto college level
        
A single electron orbits around a stationary nucleus of charge + Ze. Where Z is is a constant and e is the magnitude of the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit.
Find
  1. The value of Z.
  2. The energy required to excite the electron from the third to the fourth Bohr orbit.
  3. The wavelength of the electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
  4. The kinetic energy, potential energy, potential energy and the angular momentum of the electron in the first Bohr orbit.
  5. The radius of the first Bohr orbit.
(The ionization energy of hydrogen atom = 13.6 eV, Bohr radius = 5.3 x 10-11 metre, velocity of light = 3 x 108 m / sec. Planck’s constant = 6.6  x 1034 joules – sec)
5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							
(i) E2 = - 13.6 / 4 Z2 , E3 = - 13.6 / 9 Z2
E3 – E2 = - 13.6 Z2 (1 /9 – 1/4) = + 13.6 x 5 / 36 Z2
But E3 – E2 = 47.2 eV (Given)
∴ 13.6 x 5 / 36 Z2 = 47.2 ∴ Z = √ 47.2 x 36 / 13.6 x 5 = 5
(ii) E­4 = - 13.6 / 16 Z2
∴ E4­ – E3 = - 13.6 Z2 [ 1/ 16 – 1/9 ] = - 13.6 Z2 [ 9 – 16 / 9 x 16]
= + 13.6 x 25 x 7 / 9 x 16 = 16. 53 eV
(iii) E1 = - 13.6 / 1 x 25 = - 340 eV
∴E = E - E1 = 340 eV = 340 x 1.6 x 10-19 J [ E = 0 eV]
But E = hc / λ
∴ λ = hc / E = 6.6 x 10-34 x 3 x 108 / 340 x 10-19 x 1.6 x 10-19 m
(iv) Total Energy of 1st orbit = - 340 eV
We know that – (T.E.) = K.E. [in case of electron revolving around nucleus]
And 2T.E. = P.E.
∴ K.E. = 340 eV ; P.E. = - 680 eV
KEY CONCEPT :
Angular momentum in 1st orbit:
According to Bohr’s postulate,
mvr = nh / 2π
For n = 1,
mvr = h / 2π = 6.6 x 10-34 / 2π = 1.05 x 10-34 J-s.
(v) Radius of first Bohr orbit
r1 = 5.3 x 10-11 / Z = 5.3 x 10-11 / 5
= 1.06 x 10-11 m
5 years ago
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