﻿ In a Rutherford s α-scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 60°. What will be the number of scintillations per minute at an angle of 120°? from Physics Atoms Class 12 CBSE
In a Rutherford's α-scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 60°. What will be the number of scintillations per minute at an angle of 120°?

Given,
Number of scintillations per minute at an angle 60°, n1 = 8100 m
Number of scintillations per minute at an angle 120° , n2 =?

The scattering in the Rutherford's experiment is proportional to cot4$\frac{\mathrm{\varphi }}{2}.$

$⇒$

Therefore,

This implies,

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A single electron, orbits around a stationary nucleus of charge ze, where z is a constant and e is the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to 3rd Bohr orbit. Find,
(i) The value of z.
(ii) The energy required to excite the electron from the third to the fourth Bohr orbit.
(iii) The wavelength of electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
(iv) The kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.
(v) The radius of the first Bohr orbit.
(Ionisation energy of hydrogen atom = 13.6 eV. Bohr radius = 5.3 × 10–11 m, velocity of light = 3 × 108 m/s and Planck's constant = 6.6 × 10–34 Js)

(i) For a general hydrogen-like atom where E0 is the ionisation energy of hydrogen atom or,  (ii)  Energy required to excite the electron from the third to the fourth Bohr orbit
= 16.53 eV
(iii)  The wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity = 36.4 Å.
(iv) Kinetic energy of first Bohr orbit is numerically equal to the energy of the orbit   Potential energy of electron  Angular momentum of the electron  [For 1st Bohar arbit] (v) Radius r1 of the first Bohr orbit  369 Views

A doubly ionised lithium atom is hydrogen-like with atomic number 3:
(i) Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV.)
(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

(i) The energy difference of electron in Li++ between the first and the third orbit    Therefore, the equivalent wavelength λ is given by or, (ii) The following three spectral lines are observed due to the following transitions:
3rd to 1st orbit
3rd to 2nd orbit
2nd to 1st orbit 253 Views

Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV.

Ionization energy for hydrogen atom = 13.6 eV.
The energy of monochromatic radiation of wavelength 975 Å        ∴      Number of lines possible in the resultant spectrum = 6, as shown in Fig below. The longest wavelength will be emitted for transition from 4th orbit: to 3rd orbit with an energy.      The longest wavelength,   660 Views

The ground state energy of hydrogen atom is - 13.6 eV.
(i) What is the potential energy of an electron in the 3rd excited state?
(ii) If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted.

The energy of an electron in nth orbit is given by,

(i) For 3rd excited state, n = 4

$\therefore$ Energy at the third excited level is given by,

(ii) Required energy to jump electron to the ground state from the 3rd excited state is,

∴    Wavelength of the photon emitted is,

$⇒$

is the wavelength of the photon emitted.

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If the average life time of an excited state of hydrogen is of the order of 10–8 s, estimate how many rotation an electron makes when it is in the state n = 2 and before it suffers a transition to state n = 1. Bohr radius = 5.3 × 10–11 m.

Average lifetime of an excited state of hydrogen = 10-8 sec
Bohr radius, r = 5.3 $×$ 10-11

Velocity of electron in the nth orbit of hydrogen atom,

If,

Radius of n = 2 orbit,

$⇒$

Number of revolutions made in 1 sec

Therefore, number of revolutions made in 10-8 sec,

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