# NCERT Solutions for Class 12 Physics Chapter 12

NCERT Solutions for Class 12 Physics Chapter 12 Atom Exercises Solutions and Additional Exercises Solutions in PDF form as well as option for study online without downloading for new academic session.

Downloads Offline Apps for NCERT Sols of other subjects also for based on new CBSE Curriculum for 2020-2021.## NCERT Solutions for Class 12 Physics Chapter 12

Class: | 12 |

Subject: | Physics |

Chapter 12: | Atom |

### Class 12 Physics Chapter 12 Solutions in English

NCERT Solutions for Class 12 Physics Chapter 12 Atom in PDF form are given below to free download. Visit to Discussion Forum to ask your doubts in educational fact regarding to NIOS or CBSE boards. Download NCERT Books 2020-21 following the latest CBSE Syllabus.

### Chapter 12 Atom Solutions

- Study Online: Exercises Question Answers
- Study Online: Additional Exercises Answers
- Download Exercises in PDF form
- Download Additional Exercises in PDF
- Practice Questions for Competitive Exams
- NCERT Book Chapter 12
- NCERT Book Answers
- Revision Book Chapter 12
- Revision Book Answers
- Revision Notes 1
- Revision Notes 2
- Visit to 12th Physics Main Page

#### Important Questions for practice

1. The radius of inner most orbit of Hydrogen atom is 5.3 × 10–1m. What are the radii of n = 2 and n = 8 orbits.

2. A metal surface illuminated by 8.5 × 1014 Hz light emits electrons whose maximum energy is 0.52 eV the same surface is illuminated by 12.0 × 1014 Hz light emits elections whose maximum energy is 1.97eV. From these data find work function of the surface and value of Planck’s constant. [Work Function = 3ev]

3. State the law of radioactive disintegration. Hence define disintegration constant and half life period. Establish relation between them.

4. State Bohr’s postulates. Using these postulates, drive an expression for total energy of an electron in the nth orbit of an atom. What does negative of this energy signify?

5. The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV. What is (a) the kinetic energy, (b) the potential energy of the electron? (c) Which of the answers above would change if the choice of the zero of potential energy in changed to (i) + 0.5 eV (ii) –0.5 eV.

##### Questions from Board Papers

1. A 12.5 MeV alpha – particle approaching a gold nucleus is deflected by 180°. What is the closest distance to which it approaches the nucleus?

2. Calculate the radius of the third Bohr orbit of hydrogen atom and energy of electron in that orbit.

3. An electron and a proton are possessing same amount of K.E., which of the two have greater de-Broglie, wavelength? Justify your answer.

4. For what Kinetic energy of a neutron will the associated de Broglie wavelength be 5.6 × 10^-10m?

5. If the frequency of incident light in photoelectric experiment is doubled then does the stopping potential become double or more than double, justify?

6. What will be the distance of closest approach of a 5 MeV proton as it approaches a gold nucleus?

### Important Questions on 12th Physics Chapter 12

This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy is equal to the negative of two times of kinetic energy.

Potential energy = − 2 × (13.6) = − 27 .2 eV

The average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

The probability of scattering of α-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of α−particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.