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प्रश्न
- Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
- Calculate the orbital period in each of these levels.
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उत्तर
(a) Let v1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1.
For charge (e) of an electron, v1 is given by the relation,
`v_1 = e^2/(n_1 4pi in_0(h/(2pi))) = e^2/(2in_0h)`
Where,
e = 1.6 × 10−19 C
∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
∴ `v_1 = (1.6 xx 10^(-19))^2/(2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^-34)`
= 0.0218 × 108
= 2.18 × 106 m/s
For level n2 = 2, we can write the relation for the corresponding orbital speed as:
`v_2 = e^2/(n_2 2 in_0 h)`
= `(1.6 xx 10^(-19))^2/(2 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`
= 1.09 × 106 m/s
And, for n3 = 3, we can write the relation for the corresponding orbital speed as:
`v_3 = e^2/(n_3 2 in_0 h)`
= `(1.6 xx 10^(-19))^2/(3 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`
= 7.27 × 105 m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n = 2 and n = 3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively.
(b) Let T1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
`T_1 = (2pir_1)/v_1`
Where,
r1 = Radius of the orbit = `(n_1^2 h^2 in_0)/(pime^2)`
h = Planck’s constant = 6.62 × 10−34 Js
e = Charge on an electron = 1.6 × 10−19 C
∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
∴ `T_1 = (2pir_1)/v_1`
= `(2pi xx (1)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(2.18 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2`
= 15.27 × 10−17
= 1.527 × 10−16 s
For level n2 = 2, we can write the period as:
`T_2 = (2pir_2)/v_2`
Where, r2 = Radius of the electron in n2 = 2
`= ((n_2)^2 h^2 in_0)/(pime^2)`
∴ `T_2 = (2pir_2)/v_2`
= `(2pi xx (2)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(1.09 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2)`
= 1.22 × 10−15 s
And, for level n3 = 3, we can write the period as:
`T_3 = (2pir_3)/v_3`
Where,
r3 = Radius of the electron in n3 = 3
= `((n_3)^2h^2 in_0)/(pime^2)`
∴ `T_3 = (2pir_3)/v_3`
= `(2pi xx (3)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(7.27 xx 10^5 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19)))`
= 4.12 × 10−15 s
Hence, the orbital period in each of these levels is 1.52 × 10−16 s, 1.22 × 10−15 s and 4.12 × 10−15 s, respectively.
संबंधित प्रश्न
Draw a neat, labelled energy level diagram for H atom showing the transitions. Explain the series of spectral lines for H atom, whose fixed inner orbit numbers are 3 and 4 respectively.
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
- n = 0, l = 0, ml = 0, ms = + ½
- n = 1, l = 0, ml = 0, ms = – ½
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- n = 2, l = 1, ml = 0, ms = – ½
- n = 3, l = 3, ml = –3, ms = + ½
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b) `r_0`
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