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 = r1  ?

(b) Linear Momentum of electron =  ?

(c) Angular Momentum of electron =  ?

(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 = r1  ?

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

rn = 0.053 nm xn²

for n = 1

r1 = 0.053 nm x(1)²

r1 = 0.053 nm x1

r1 = 0.053 nm  ------------------Ans.1



(b) Linear Momentum of electron = P  ?

Linear Momentum at any energy level 'n' formula is 

P = mvn 

at n = 1

P = mv1 

Here we have to find first the speed of electron v1 So,

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

 vn = 2πke²nh


by putting the corresponding values 

v₁ = 23.1416910Nm²C²(1.610¹C)²16.6210³m²kgs¹

[N = kg m s⁻²]

v₁ = 56.548810kgms²m²C²2.5610³C²6.6210³m²kgs¹

v₁ = 144.76492810²kgs²m³C²6.6210³m²kgs¹

v₁ = 21.868x 10⁵ m s⁻¹

or

v₁ = 2.187x 10 m s⁻¹

Now 

P = mv1 = 9.11 x 10⁻³¹ kg x 2.187x 10 m s⁻¹

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



(c) Angular Momentum of electron =  ?

We know that 

L = mvr

at n=1

mvr

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

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


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

K.E 12mv₁²

K.E12x 9.11 x 10⁻³¹ kg x (2.187x 10 m s⁻¹)²

K.E 12 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 2.1810¹1.610¹ eV

K.E = 13.6  eV ----------------Ans. 4


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

Since at n=1

P.E   - ke²r

by putting values

P.E   - 910Nm²C²(1.610¹C)²0.05310m

P.E   - 4.35 x 10⁻¹⁸ J

To express K.E. in Electron Volts we have

P.E -4.3510¹1.610¹ eV

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


(f) Total Energy at (n = 1) = E ?

As

E  K.E1  + P.E1 

E  13.6  eV + (-27.2  eV  )

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


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