Quantum Mechanics: Important Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

 

1. An electron is accelerated by a potential difference of 150 V. What is the wave length of that electron wave?

Solution

Given: Accelerated voltage of the electron (V) = 150 V

Formula: de Broglie wavelength λ = h / √(2meV)

 λ = [ 6.625 × 10^‒34 ] / √[2 × 9.11 × 10^‒31 × 1.6 × 10^‒19 × 150]

(or) λ = [ 6.625 × 10^‒34 ] / √[4.3728 × 10^‒47]

(or) λ =  6.625 × 10^‒34  / 6.6127 × 10^‒24

λ= 1.0018 × 10⁻¹⁰ metres

. The de‒Broglie wavelength λ=1Å

 

2. An electron at rest is accelerated through a potential of 5000 V. Calculate the de Broglie wavelength of matter wave associated with it.

Solution

Given: V = 5000 V

Formula: de‒Broglie wavelength λ = h / √(2meV)

λ = [ 6.625 × 10^‒34 ] / √[2 × 9.11 × 10^‒31 × 1.6 × 10^‒19 × 5000]

(or) λ = [ 6.625 × 10^‒34 ] / √[1.4576 × 10^‒45]

(or) λ =  6.625×10^‒34  / 3.8178×10^-23

λ= 0.17353 × 10^‒10 m (or) 0.17353 Å

∴ de‒Broglie wavelength of electron λ= 0.17353 Å

 

3. Calculate the de‒Broglie wave length of an electron of energy 100 eV

Solution

Given: Energy of electron (E) = 100 eV

E = 100 × 1.6 × 10^‒19 Joules

E = 1.6 × 10^‒17 Joules

Formula: We know that de‒Broglie wavelength λ= h / √(2mE)

Substituting the given values, we have

λ = [ 6.625 × 10^‒34 ] / √[2 × 9.11 × 10^‒31 × 1.6 × 10^‒17]

λ =  6.625 × 10^‒34  / 5.3993× 10^-47

λ= 1.227× 10^‒10 m (or) 1.227Å

:. de‒Broglie wavelength λ= 1.227 Å

 

4. Calculate the de‒Broglie wave length associated with a proton moving with a velocity equal to 1/20 th of the velocity of light.

Mass of proton = 1.675 × 10^‒27 kg

Solution

Given: Mass of the proton m=1.675 × 10^‒27 kg

Velocity of proton v = 1/20 × velocity of light

v= 1/20 × 3 × 10⁸

Velocity (v) = 15 × 10⁶ m/s

Formula: de‒Broglie wavelength λ = h / mv

Substituting the given values

we have λ = [ 6.625 × 10⁻³⁴ ] /  [ 1.675 × 10⁻²⁷ × 1.5 × 10⁷ ]

= 2.6447 × 10^‒14 m

 ∴ de‒Broglie wavelength λ= 2.6447 × 10^‒14 m

 

5. A neutron of mass 1.675 × 10^‒27 kg is moving with a kinetic energy 10 KeV. Calculate the de‒Broglie wavelength associated with it.

Solution

Given: Energy of neutron = 10 KeV

E=10× 10³ × 1.6 × 10^‒19 Joules

E=1.6×10^‒15 Joules

Formula: de‒Broglie wavelength λ= h / √(2mE)

λ = [ 6.625 × 10^‒34 ] / √[2 × 1.675 × 10^‒27 × 1.6 × 10^‒15]

λ =  6.625 × 10^‒34  / 2.3156× 10^-21

λ= 2.861 × 10^‒13 m

∴ de‒Broglie wavelength of neutron = 2.861 × 10^‒13 m

 

6. Calculate the minimum energy an electron can possess in an infinitely deep potential well of width 4 nm.

Solution

Given: l= 4 nm = 4 × 10^‒9m

We know for minimum energy n = 1

Formula: Energy of an electron in an infinitely deep potential well

 E = n²h² / 8ml²

E= [ l² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (4 × 10^‒9)² ]

(or) E= 4.3891×10^‒67 / 1.16608×10^‒46

E = 3.7639 × 10^‒21 J

(or) E = 3.7639×10^‒21 / 1.6×10^‒19   eV

 E=0.02352 eV

 Minimum energy of an electron (E) = 0.0235 eV

 

7. An electron is trapped in a one dimensional box of length 0.1 nm. Calculate the energy required to excite the electron from its ground state to the fifth excited state.

Solution

Given: l = 0.1 nm = 0.1 × 10^‒9 m

Formula: Energy of an electron in one dimensional box E = n²h² / 8ml²

We know for ground state n=1

E_g= [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.1 × 10^‒9)² ]

(or) E_g= 4.3891×10^‒67 / 7.288×10^‒50

(or) E_g= 6.0223×10^‒18  J

(or) E_g= 6.0223×10^‒18 / 1.6×10^‒19      eV

E_g = 37.64 eV

For fifth excited state n=6

E= [ 6² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.1 × 10^‒9)² ]

 = 1.58007×10^‒65 / 7.288×10^‒50

E_e = 2.16805 × 10^‒16 J (or)

E_e = 2.16805×10^‒16 / 1.6×10^‒19 eV

E_e=1355.03 eV

The energy required to excite the electron from its ground state to the fifth excited state is E=E_e‒E_g

E = 1355.03‒37.64 eV

(or) E=1317.39 eV

Energy required to excite an electron from fifth excited state to ground state = 1317.39 eV

 

8. Find the energy of an electron moving in one dimension in an infinitely high potential box of width 0.1 nm.

Solution

Given: Length (or) width of one dimension box l = 10 nm = 0.1×0.1^‒9 m

Formula: The energy of an electron is E =  n²h² / 8ml²

For Lowest energy n = 1.

E =  n²h² / 8ml²

E_g= [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.1 × 10^‒9)² ]

(or) E_g= 4.3891×10^‒67 / 7.288×10^‒50

(or) E_g= 6.0223×10^‒18  J

(or) E_g= 6.0223×10^‒18 / 1.6×10^‒19      eV

E_g = 37.639 eV

Energy of the electron E = 37.639 eV

 

9. Calculate the least energy that an electron can possess in a one‒dimensinal potential box of width 0.5 nm and infinite height. (m=9.11 × 10^‒31 kg)

Solution

Formula: The energy of an electron is E = n²h² / 8ml²

The energy of an electron in a one dimensional potential box is given by E =  n²h² / 8ml²

For Lowest energy n = 1.

E =  n²h² / 8ml²

E₁= [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.5 × 10^‒9)² ]

(or) E₁= 4.3891×10^‒67 / 1.822×10^‒48

(or) E₁= 2.405 × 10^‒19 Joules

(or) E₁= 2.405×10^‒19  / 1.6×10^‒19      eV

(or) E₁ = 1.5055 eV

The least energy of the electron (E₁) = 1.5055 eV

 

10. Calculate the de‒Broglie wavelength of an electron which has been accelerated from rest on application of potential of 400 volts.

Solution

Given Data:

V = 400 Volts

Formula:

(i) de‒Broglie wavelength λ = h / √(2meV)

 = 6.625×10^‒34 / √[2×9.11×10^‒31× 1.6×10^‒19 × 400]

λ= 6.625×10^‒34 × 1.07985×10^‒23

λ= 6.135 × 10^‒11

(or) λ=0.6138 Å

The de‒Broglie wavelength = 0.6135 Å

 

11. The de‒Broglie wavelength of an electron is 1.226 Å. What is the value of accelerating potential?

Solution

Given Data:

 λ= 1.226 Å (or) 1.226 × 10^‒10 m

Formula:

The de‒Broglie wave length λ = h / √(2meV)

λ² = h² / 2meV

V = h² / 2meλ²

Accelerating Potential

V = h² / 2meλ²

V = [6.625×10⁻³⁴]² / [2× 9.11×10^‒31 × 1.6×10^‒19 × (1.226 × 10^‒10)²]

V= 4.3891×10^‒67 / 4.3818×10^‒69

 V= 100.16 Volts

Accelerating Voltage = 100 Volts

 

12. Calculate the equivalent wavlength of electron moving with a velocity of 3 × 10⁷ m/s.

Solution

Given Data:

Velocity (V) = 3 × 10⁷ m/s

Formula:

de‒Broglie wavelength λ = h / mv

 λ = [6.625×10^‒34] / [9.11 × 10⁻³¹ × 3 × 10⁷]

 λ = 6.625×10^‒34 / 2.733×10^‒23

 λ = 2.424 × 10^‒11 meters

or

λ=0.2424 Å

The de‒Broglie wavelength of electron = 0.2424 Å

 

13. An electron is confined to a one dimensional box of side 10^‒10 m. Obtain the first four Eigen values of the electron.

Solution

Formula: Energy eigen value E_n =  n²h² / 8ml²

1st Eigen value E₁ = [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (10^‒10)² ]

(or) E₁= 4.38906×10^‒67 / 7.288×10^‒50

(or) E₁= 6.0223×10^‒18  J

(or) E₁= 6.0223×10^‒18 / 1.6×10^‒19      eV

E₁ = 37.6375 eV

2^nd Eigen value   E₂=2² E₁ = 2.4089 × 10^‒17 J (or) 150 eV

3^rd Eigen value   E₃=3² E₁ = 5.4198 × 10^‒17 J (or) 338 eV

4^th Eigen value   E₄=4² E₁ = 9.6352 × 10^‒17 J (or) 602 eV

 

14. Calculate the energy in eV of a photon of wavlength 1.2 Å (Planck's constant = 6.62 × 10^‒34 Js; speed of light = 3 × 10⁸ m/s)

Solution

Given data:

λ=1.2 × 10^‒10m

h=6.62 × 10^‒34 Js

c=3×10⁸ m/s

Formula:

E = hv

E = hc/λ

E = 6.625×10^‒34×3×10⁸  /  1.2×10^‒10

E = 1.986×10^‒25  /  1.2×10^‒10

E= 1.655 × 10⁻¹⁵ Joules

E= 1.655×10⁻¹⁵ / 1.6×10^‒19      eV

E=10343.75 eV

Energy of the photon = 10343.75 eV

 

15. Calculate the energy of an electron moving in one‒dimension in an infinitely high potential box of width 0.3 nm, if the mass of the electron is 9.11 × 10^‒31 kg and Planck's constant is 6.625 × 10^‒31 Js.

Solution

Given data:

 l = 0.3 × 10^‒9 m

 m = 9.11 × 10^‒31 kg

 h = 6.625 × 10^‒34 Js

 n = 1

Formula:

E =  n²h² / 8ml²

E = [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.3×10^‒9)² ]

(or) E= 4.38906×10^‒67 / 6.5592×10^‒49

(or) E= 6.6914×10^‒19  J

(or) E= 6.6914×10^‒19 / 1.6×10^‒19      eV

E₁ = 4.182 eV

Energy of electron moving in one dimensional box = 4.182 eV

 

16. Calculate de Broglie wavelength of an electron accelerated to a potential of 100 Volts.

Solution

Given data:

V=100 volts

Formula: de‒Broglie wavelength λ = h / √(2meV)

λ = h / √(2meV)

 = 6.625×10^‒34 / √[2×9.11×10^‒31× 1.6×10^‒19 × 100]

λ= 6.625×10^‒34 × 5.399×10^‒24

λ= 1.227 × 10^‒10 m

(or) λ=1.227 Å

The de‒Broglie wavelength of electron = 1.227 Å

 

17. Find the lowest energy of an electron confined in a box of length 0.2 nm.

Solution

Given data:

 l = 0.2 × 10^‒9 m

n=1

h = 6.625 × 10^‒34 Js

Formula: The energy of an electron is E = n²h² / 8ml²

For Lowest energy n = 1.

E =  n²h² / 8ml²

E= [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (0.2 × 10^‒9)² ]

E= 4.3891×10^‒67 / 2.9152×10^‒49

E= 1.5056 × 10^‒18 Joules

E= 1.5056×10^‒18  / 1.6×10^‒19      eV

(or) E = 9.41 eV

Lowest energy of an electron = 9.41 eV

 

ADDITIONAL SOLVED PROBLEMS

 

1. Calculate the number of photons emitted by a 100 watts sodium vapour lamp. [Give λ= 5893 Å].

Solution

Formula: Energy = hv = hc/λ

E = hc/λ

E = 6.625×10^‒34×3×10⁸  /  5893×10^‒10

E= 3.3726 × 10⁻¹⁹ Joules

∴ Number of photons emitted = Power / Energy

= 100J/S / 3.3726×10^‒19joules

= 2.965 × 10²⁰ per second

Number of photons emitted = 2.965 × 10²⁰

 

2. Calculate the de‒Broglie wavelength of an electron accelerated to a potential of 2 KV.

Solution

Formula: de‒Broglie wavelength λ = h / √(2meV)

λ = h / √(2meV)

 = 6.625×10^‒34 / √[2×9.11×10^‒31× 1.6×10^‒19 × 2× 10³]

 (or) λ = 0.2744 Å

The de‒Broglie wavelength λ = 0.2744 Å

 

3. Calculate the de‒Broglie wavelength corresponding to the root mean square velocity of hydrogen molecules at 27°C.

Solution

Formula: de‒Broglie wavelength λ= h / √(3mK_BT)

Mass of hydrogen = mass of proton = 1.678 × 10^‒27 kg

Boltzmann constant K_B = 1.38 × 10^‒23

Temperature (T) = 300 K

λ = 6.625×10^‒34 / [3×1.678 × 10^‒27 × 1.38 × 10^‒23 × 300]

λ = 1.451 Å

de‒Broglie wavelength (λ) = 1.451 Ä

 

4. Calculate the energy in eV of an electron wave of wavelength 3× 10^‒2 m.

Solution

Formula: Energy E = h² / 2mλ²

E = h² / 2mλ²

E = [6.625×10⁻³⁴]² / [2× 9.11×10^‒31 × (3 × 10^‒2)²]

E = 2.6765 × 10^‒34 Joules

E = 2.6765×10^‒34 / 1.6×10^‒19      eV

E = 1.6728 eV

∴ Energy of the electron 'E' = 1.6728 eV

 

5. Calculate the lowest energy of the system containing two electrons confined to a box of length 1 Å.

Solution

Formula: Energy of the system having two electrons E = 2( n²h² / 8ml² )

E = 2( n²h² / 8ml² )

E = 2 [ 1² × (6.625 × 10^‒34)² ] / [ 8 × 9.11 × 10^‒31 × (1 × 10^‒10)² ]

E= 1.2044× 10^‒17 Joules

E= 1.2044×10^‒17  / 1.6×10^‒19      eV

(or) E = 75.275 eV

Energy of the system having two electrons = 75.275 eV

 

6. Calculate the reflection and transmission co‒efficient of an electron of energy 15 eV incident on a potential barrier of height 10 eV.

Solution

E = 15 eV

V = 10 eV

Given data

Formula

(i) Reflection co‒efficient R = [ √E‒√(E‒V)  /  [√E+√(E –V) ]²

∴ R = 0.072

(ii) Transmission co‒efficient T = 1‒R

∴ T = 1 − 0.072

∴ T=0.928

The Reflection co‒efficient R= 0.072

& The transmission co‒efficient T=0.928

 

7. Calculate the reflection co‒efficient for a particle of kinetic energy 9 eV incident on a potential step of height 5 eV.

Solution

Given data

E = 9 eV

V = 5 eV

Formula

Reflection co‒efficient R = [ √E‒√(E‒V)  /  [√E+√(E –V) ]²

∴ R = [ 1/5 ]²

∴ R = 1/25

(or) R= 0.04

Reflection co‒efficient R= 0.04

 

8. Show that the sum of reflection and transmission co‒efficient is one for a particle scattered by a potential step.

Solution

R+T=1

 (ii) For E<V

Here for E<V, transmission will not be there

∴ T=0

R+T=1+0

R+T=1

 

ASSIGNMENT PROBLEMS

 

1. Calculate the energy of the photon having the same momentum as that of 10 MeV proton. (Ans: E=13.69 × 10⁷ eV)

 

2. What is the energy of neutrons whose de‒Broglie wavelength is 1 Å. [Given: Mass of neutron = 1.6747 × 10^‒27 kg] (Ans: E=1.3104 × 10^‒20 J)

 

3. Find the de‒Broglie wavelength for an electron, accelerated through a potential difference of 28.8 V. (Ans: λ = 2.4 Å)

 

4. What is the momentum of the proton having kinetic energy of 1 BeV. (Ans: p =7.31 × 10^‒19 kg ms^‒1)

 

5. Calculate the lowest energy of the system consisting of three electrons in a one dimension potential box of length 1 Å. (Ans: E=225.93 eV)

 

6. Calculate the probability of transmission that a 1 eV electron will penetrate a potential barrier of 4 eV, when the width of the barrier is 2 Å. (Ans: T=0.084 (or) T=8.4%)

 

7. A particle of energy E is incident on a potential step of barrier height V. What should be the ratio E/V so that the probability of transmission is 50%? (Ans: E/V=1.03)