Solved Numerical Problem No. 20.1 ( Unit-20:Atomic Spectra); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

A hydrogen atom is in its ground state (n = 1). Using Bohr’s theory, calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angular momentum of the electron (d) the kinetic energy (e) the potential energy, and (f)the total energy. [Ans: (a) 0.529 ｘ 10⁻¹⁰ m (b) 1.99 ｘ 10⁻²⁴ kg m s⁻¹ (c) 1.05 ｘ 10⁻³⁴ kg m² s⁻¹ (d ) 13.6 eV (e) - 27.2 eV (0 -13.6 eV)

Data Given:

Hydrogen Atom

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

∴ Charge on the electron = e = 1.6 ｘ 10⁻¹⁹ C

∴ Coulomb's Constant = K = 9 ｘ 10⁹ N m² C⁻²

To Find:

(a) Radius of the 5th energy level =

r_{1}

$\r_1$ = ?

(b) Linear Momentum of electron = P = ?

(d) Kinetic Energy of electron = K.E = ?

(e) Potential Energy of electron = P.E = ?

(f) Total Energy at (n = 1) = E = ?

Solution:

(a) Radius of the 5th energy level =

r_{1}

$\r_1$ = ?

The general formula for the radius of quantized orbit for a Hydrogen atom is

r_{n}

$\r_n$ = 0.053 nm ｘn²

for n = 1

r_{1}

$\r_1$ = 0.053 nm ｘ(1)²

r_{1}

$\r_1$ = 0.053 nm ｘ1

r_{1}

$\r_1$ = 0.053 nm ------------------Ans.1

(b) Linear Momentum of electron = P = ?

Linear Momentum at any energy level 'n' formula is

P = m

v_{n}

$\v_n$

at n = 1

P = m

v_{1}

$\v_1$

Here we have to find first the speed of electron

v_{1}

$\v_1$ So,

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

v_{n}

$\v_n$ =

\frac{2 π k e ²}{n h}

$\frac{2πke²}{nh}$

by putting the corresponding values

v₁ =

\frac{2 ｘ 3.1416 ｘ 9 ｘ 10 ⁹ N m ² C ⁻ ² ｘ (1.6 ｘ 10 ⁻ ¹ ⁹ C) ²}{1 ｘ 6.62 ｘ 10 ⁻ ³ ⁴ m ² k g s ⁻ ¹}

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

[N = kg m s⁻²]

v₁ =

\frac{56.5488 ｘ 10 ⁹ k g m s ⁻ ² m ² C ⁻ ² ｘ 2.56 ｘ 10 ⁻ ³ ⁸ C ²}{6.62 ｘ 10 ⁻ ³ ⁴ m ² k g s ⁻ ¹}

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

v₁ =

\frac{144.764928 ｘ 10 ⁻ ² ⁹ k g s ⁻ ² m ³ C ⁻ ²}{6.62 ｘ 10 ⁻ ³ ⁴ m ² k g s ⁻ ¹}

$\frac{144.764928 ｘ 10⁻²⁹ kg s⁻² m³ C⁻²}{ 6.62 ｘ 10⁻³⁴ m² kg s⁻¹}$

v₁ = 21.868ｘ 10⁵ m s⁻¹

v₁ = 2.187ｘ 10⁶ m s⁻¹

Now

P = m

v_{1}

$\v_1$ = 9.11 ｘ 10⁻³¹ kg ｘ 2.187ｘ 10⁶ m s⁻¹

P = 1.99 ｘ 10⁻²⁴ kg m s⁻¹ -------------Ans.2

We know that

L = mvr

at n=1

L = mv₁r₁

L = 9.11 ｘ 10⁻³¹ kg ｘ 2.187ｘ 10⁶ m s⁻¹ ｘ 0.053 ｘ 10⁻⁹ m

L = 1.0547 ｘ 10⁻³⁴ kg m² s⁻¹ ---------------Ans.3

(d) Kinetic Energy of electron = K.E = ?

K.E =

\frac{1}{2}

$\frac {1}{2}$ mv₁²

K.E=

\frac{1}{2}

$\frac {1}{2}$ ｘ 9.11 ｘ 10⁻³¹ kg ｘ (2.187ｘ 10⁶ m s⁻¹)²

K.E =

\frac{1}{2}

$\frac {1}{2}$ ｘ 9.11 ｘ 10⁻³¹ kg ｘ 4.782969ｘ 10¹² m² s⁻²

K.E = 2.18 ｘ 10⁻¹⁸ J

To express K.E. in Electron Volts we have

K.E =

\frac{2.18 ｘ 10 ⁻ ¹ ⁸}{1.6 ｘ 10 ⁻ ¹ ⁹}

$\frac {2.18 ｘ 10⁻¹⁸ }{1.6 ｘ 10⁻¹⁹ }$ eV

K.E = 13.6 eV ----------------Ans. 4

(f) Potential Energy at (n = 1) = P.E = ?

Since at n=1

P.E = -

\frac{k e ²}{r ₁}

$\frac{ke²}{r₁}$

by putting values

P.E = -

\frac{9 ｘ 10 ⁹ N m ² C ⁻ ² ｘ (1.6 ｘ 10 ⁻ ¹ ⁹ C) ²}{0.053 ｘ 10 ⁻ ⁹ m}

$\frac{9 ｘ 10⁹ N m² C⁻² ｘ (1.6 ｘ 10⁻¹⁹ C)²}{0.053 ｘ 10⁻⁹ m}$

P.E = - 4.35 ｘ 10⁻¹⁸ J

To express K.E. in Electron Volts we have

P.E =

\frac{- 4.35 ｘ 10 ⁻ ¹ ⁸}{1.6 ｘ 10 ⁻ ¹ ⁹}

$\frac {- 4.35 ｘ 10⁻¹⁸ }{1.6 ｘ 10⁻¹⁹ }$ eV

P.E = -27.2 eV ----------------Ans. 5

(f) Total Energy at (n = 1) = E = ?

E =

K . E_{1}

$\K.E_1$ +

P . E_{1}

$\P.E_1$

E = 13.6 eV + (-27.2 eV )

E = - 13.6 eV --------------------Ans.6

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