Solved Numerical Problem No.5 ( Unit-19:Atomic Spectra); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

An electron is in the first Bohr orbit of the hydrogen atom. (a) Find the speed of an electron, and (b) the time required for an electron to circle the nucleus. (Answer: (a) 2.19 ｘ 10⁶ m/s (b) 1.52 ｘ 10⁻¹⁶ s) 

Given:

Atom: Hydrogen

Energy level n = 1

∴ Plank's Constant = h =  6.62 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Mass of the electron = m = 9.11 ｘ 10⁻³¹ kg

∴ Radius of 1st Bohr Orbit = r = 0.53 ｘ 10⁻¹⁰ m

To Find:

(a) Speed of electron in the first orbit = v₁ =  ?

(b) Time period = T₁ =  ?

Solution:

(a) Speed of electron in the first orbit = v₁ =  ?

The  general formula for finding the velocity of an electron in any orbit is 

 `\v_n` = `\frac{2πke²}{nh}`

by putting the corresponding values 

v₁ = `\frac{2 ｘ 3.1416 ｘ 9 ｘ 10⁹ N m² C⁻² ｘ (1.6 ｘ 10⁻¹⁹ C)²}{1 ｘ 6.62 ｘ 10⁻³⁴ m² kg s⁻¹}`

[N = kg m s⁻²]

v₁ = `\frac{56.5488ｘ 10⁹ kg m s⁻² m² C⁻² ｘ 2.56 ｘ 10⁻³⁸ C²}{ 6.62 ｘ 10⁻³⁴ m² kg s⁻¹}`

v₁ = `\frac{144.764928 ｘ 10⁻²⁹ kg s⁻² m³ }{ 6.62 ｘ 10⁻³⁴ m² kg s⁻¹}`

v₁ = 21.868ｘ 10⁵ m s⁻¹

or

v₁ = 2.187ｘ 10⁶ m s⁻¹ ---------------Ans.1

(b) Time period = T₁ =  ?

We know that 

S = v t

or 

t = `\frac {S}{v}`

where s = circumference of the orbit = 2πr

t = `\frac {2πr}{v}`

for n = 1 , t = T

T₁ = `\frac {2πr₁}{v₁}`

by putting values

T₁ = `\frac {2 ｘ 3.1416 ｘ 0.53 ｘ 10⁻¹⁰ m}{2.187ｘ 10⁶ m s⁻¹}`

T₁ = `\frac {3.330096 ｘ 10⁻¹⁰ m}{2.187ｘ 10⁶ m s⁻¹}`

T₁ = 1.523 ｘ 10⁻¹⁶ s⁻¹ ------------Ans.2

Similar Questions:

The orbital electron of a hydrogen atom moves with a speed of 5.456 ｘ10⁵ m s⁻¹ (a) find the value of the quantum number 'n' associated with this electron,(b) calculate the radius of this orbit, and (c) the energy of the electron in this orbit? (Ans: n = 4. r₄ = 0.846 nm: E₄ = -0.85 eV)

In the fifth energy level determine the radius and energy of electron for hydrogen atom? (Answer: 1.323 nm, -8.72 x 10⁻²⁰ J or -0.545 eV) 

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