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 = P = ?
(c) Angular Momentum of electron = L = ?
(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 = 2x3.1416x9x10⁹Nm²C⁻²x(1.6x10⁻¹⁹C)²1x6.62x10⁻³⁴m²kgs⁻¹
[N = kg m s⁻²]
v₁ = 56.5488x10⁹kgms⁻²m²C⁻²x2.56x10⁻³⁸C²6.62x10⁻³⁴m²kgs⁻¹
v₁ = 144.764928x10⁻²⁹kgs⁻²m³C⁻²6.62x10⁻³⁴m²kgs⁻¹
v₁ = 21.868x 10⁵ m s⁻¹
or
v₁ = 2.187x 10⁶ m s⁻¹
Now
P = mv1 = 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
(c) Angular Momentum of electron = L = ?
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 = 12mv₁²
K.E= 12x 9.11 x 10⁻³¹ kg x (2.187x 10⁶ 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 = 2.18x10⁻¹⁸1.6x10⁻¹⁹ eV
K.E = 13.6 eV ----------------Ans. 4
(f) Potential Energy at (n = 1) = P.E = ?
Since at n=1
P.E = - ke²r₁
by putting values
P.E = - 9x10⁹Nm²C⁻²x(1.6x10⁻¹⁹C)²0.053x10⁻⁹m
P.E = - 4.35 x 10⁻¹⁸ J
To express K.E. in Electron Volts we have
P.E = -4.35x10⁻¹⁸1.6x10⁻¹⁹ eV
P.E = -27.2 eV ----------------Ans. 5
As
E = K.E1 + P.E1
E = 13.6 eV + (-27.2 eV )
E = - 13.6 eV --------------------Ans.6
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