Atoms

Question 1

alpha particle scattering experiment - observations

Answer 1

1. Most of the alpha particles pass through without appreciable deflection.
2. Number of alpha particles scattered decreases sharply with angle of scattering
3. Only very few (1 out of 10000) suffer 180-degree deflection.

Question 2

alpha particle scattering experiment - conclusions

Answer 2

1. Most part of the atom is empty space
2. Most mass of the atom is concentrated in a small space.
3. The positive charge of the atom along with the mass is concentrated into a small space called nucleus.

Question 3

distance of closest approach (d) =

Answer 3

(1/4πε)×(2Ze^2/KE)

Question 4

Define Impact Parameter.

Answer 4

Impact parameter is defined as the perpendicular distance between the initial velocity vector of the alpha particle and the line passing through the center of nucleus.

Question 5

Expressions for potential, kinetic and total energy based on rutherford’s model.

Answer 5

PE = (1/4πε)×(-e^2/r)
KE = (1/8πε)×(e^2/r)
Total energy = -(1/8πε)×(e^2/r)

Question 6

Two limitations of Rutherford’s model

Answer 6

1. Could not explain the stability of atom as electrons revolving around the nucleus are supposed to radiate energy in the form of electromagnetic waves, as the energy decreases, the electrons are supposed to spiral into the nucleus and collapse but practically atoms are stable.
2. Could not explain the line spectra of hydrogen atoms (could only explain continuous spectra). Since the electron spiral with continuously increasing frequency, it is supposed to emit only continuous spectrum of radiations.

Question 7

The 1st postulate of Bohr’s model of an atom

Answer 7

Electrons around the nucleus can continue revolving without radiating energy in certain specified orbits known as non-radiating orbits or stable orbits.

Question 8

The 2nd postulate of Bohr’s model of an atom

Answer 8

Electrons whose angular momentum is an integral multiple of h/2π can revolve without radiating energy.

Question 9

The 3rd postulate of Bohr’s model of an atom

Answer 9

Electrons deexciting from higher energy level to lower energy level radiate energy equal to the difference in the energy levels of initial and final orbits.

Question 10

Expressions for potential, kinetic and total energy according to Bohr’s model

Answer 10

PE =  (1/4πε)×(-me^4/n^2×h^2)
KE =  (1/8πε)×(me^e4/n^2×h^2)
TE =  -(1/8πε)×(me^4/n^2×h^2)

Question 11

Expression for wavelength of photon emitted due to deexcitation of electron

Answer 11

1/λ = R((1/nf^2) - (1/ni^2))

Question 12

for an electron to show stationary wave pattern, the circumference of the orbit =

Answer 12

nλ (where λ is the wavelength of electron in nth orbit/debroglie’s wavelength)

Question 13

Debroglie’s wavelength =

Answer 13

h/mv

Question 14

Limitations of Bohr’s Model

Answer 14

1. Bohr failed to explain the splitting of spectral lines in magnetic field (Zeimann effect) and electric field (Stark effect).
2. Bohr’s model was applicable only for hydrogen atom.
3. Bohr failed to explain the relative intensity of different spectral lines.

Question 15

number of spectral lines emitted when an electron undergo deexcitation from level n =

Answer 15

n(n-1)/ 2

Question 16

hydrogen spectrum, n = 1

Answer 16

Lymann series
UV region
E1 = -13.6 eV

Question 17

hydrogen spectrum, n = 2

Answer 17

Balmer series
Visible region
E2 = -3.4 eV

Question 18

hydrogen spectrum, n = 3

Answer 18

Paschen series
Infrared region
E3 = -1.51 eV

Question 19

hydrogen spectrum, n = 4

Answer 19

Brackett series
Infrared region
E4 = -0.85 eV

Question 20

hydrogen spectrum, n = 5

Answer 20

Pfund series
Infrared region
E5 = -0.54 eV