Science (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Physics

The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm⁻¹ at its surface in the downward direction, corresponding to a surface charge density = −10⁻⁹ C m⁻². Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

The energy associated with the first orbit in the hydrogen atom is - 2.18 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit?

Calculate the radius of Bohr’s fifth orbit for hydrogen atom

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10^–11 ergs.

Explain, giving reasons, which of the following sets of quantum numbers are not possible.

1. n = 0, l = 0, ml = 0, ms = + ½
2. n = 1, l = 0, ml = 0, ms = – ½
3. n = 1, l = 1, ml = 0, ms = + ½
4. n = 2, l = 1, ml = 0, ms = – ½
5. n = 3, l = 3, ml = –3, ms = + ½
6. n = 3, l = 1, ml = 0, ms = + ½

How many electrons in an atom may have the following quantum numbers?

n = 4, `m_s =  -1/2`

How many electrons in an atom may have the following quantum numbers?

n = 3, l = 0

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Calculate the energy required for the process 

\[\ce{He^+_{(g)} -> He^{2+}_{(g)} + e^-}\]

The ionization energy for the H atom in the ground state is 2.18 ×10^–18 J atom^–1

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has a duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 10¹⁵, calculate the energy of the source.

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 10⁷ ms^–1, calculate the energy with which it is bound to the nucleus.

If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ ms^-1, calculate the de Broglie wavelength associated with it.

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?

1. Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
2. Calculate the orbital period in each of these levels.

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n = 2 and n = 3 orbits?

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg)