Question 2.49:
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.
Answer:
Frequency of radiation (ν),
ν = 5.0 × 108 s–1
Energy (E) of source = Nhν
Where,
N = number of photons emitted
h = Planck’s constant
ν = frequency of radiation
Substituting the values in the given expression of (E):
E = (2.5 × 1015) (6.626 × 10–34 Js) (5.0 × 108 s–1)
E = 8.282 × 10–10 J
Hence, the energy of the source (E) is 8.282 × 10–10 J.
Question 2.50:
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.
Answer:
For λ1 = 589 nm
Frequency of transition
Frequency of transition (ν1) = 5.093 × 1014 s–1
Similarly, for λ2 = 589.6 nm
Frequency of transition
Frequency of transition (ν2) = 5.088 × 1014 s–1
Energy difference (ΔE) between excited states = E1 – E2
Where,
E2 = energy associated with λ2
E1 = energy associated with λ1
ΔE = hν1 – hν2
= h(ν1 – ν2)
= (6.626 × 10–34 Js) (5.093 × 1014 – 5.088 × 1014)s–1
= (6.626 × 10–34 J) (5.0 × 10–3 × 1014)
ΔE = 3.31 × 10–22 J
Question 2.51:
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer:
It is given that the work function (W0) for caesium atom is 1.9 eV.
(a) From the expression, , we get:
Where,
λ0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of (λ0):
6.53 × 10–7 m
Hence, the threshold wavelength is 653 nm.
(b) From the expression, , we get:
Where,
ν0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν0:
(1 eV = 1.602 × 10–19 J)
ν0 = 4.593 × 1014 s–1
Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014 s–1.
(c) According to the Question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν0)
= 9.3149 × 10–20 J
Kinetic energy of the ejected photoelectron = 9.3149 × 10–20J
Since K.E
v = 4.52 × 105 ms–1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105 ms–1.
Question 2.52:
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
λ (nm) |
500 |
450 |
400 |
v × 10–5 (cm s–1) |
2.55 |
4.35 |
5.35 |
(a) Assuming the threshold wavelength to be , the kinetic energy of the radiation is given as:Answer:
Three different equalities can be formed by the given value as:
Similarly,
Dividing equation (3) by equation (1):
Threshold wavelength = 540 nm
Note: part (b) of the Question is not done due to the incorrect values of velocity given in the Question.
Question 2.53:
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:
From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W0 + K.E
⇒ W0 = E – K.E
Energy of incident photon (E)
Where,
c = velocity of radiation
h = Planck’s constant
λ = wavelength of radiation
Substituting the values in the given expression of E:
E = 4.83 eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,
K.E = 0.35 V
K.E = 0.35 eV
Work function, W0 = E – K.E
= 4.83 eV – 0.35 eV
= 4.48 eV
Question 2.54:
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.
Answer:
Energy of incident photon (E) is given by,
Energy of the electron ejected (K.E)
= 10.2480 × 10–17 J
= 1.025 × 10–16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252 × 10–16 J – 1.025 × 10–16 J
= 12.227 × 10–16 J
Question 2.55:
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:
Wavelength of transition = 1285 nm
= 1285 × 10–9 m (Given)
(Given)
Since
ν = 2.33 × 1014 s–1
Substituting the value of ν in the given expression,
n = 4.98
n ≈ 5
Hence, for the transition to be observed at 1285 nm, n = 5.
The spectrum lies in the infra-red region.
Question 2.56:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
The radius of the nth orbit of hydrogen-like particles is given by,
For radius (r1) = 1.3225 nm
= 1.32225 × 10–9 m
= 1322.25 × 10–12 m
= 1322.25 pm
Similarly,
⇒ n1 = 5 and n2 = 2
Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series.
Wave number for the transition is given by,
1.097 × 107 m–1
= 2.303 × 106 m–1
Wavelength (λ) associated with the emission transition is given by,
= 0.434 ×10–6 m
λ = 434 nm
​This transition belongs to Balmer series and comes in the visible region of the spectrum.
Question 2.57:
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.
Answer:
From de Broglie’s equation,
= 4.55 × 10–10 m
λ = 455 pm
de Broglie’s wavelength associated with the electron is 455 pm.
Question 2.58:
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:
From de Broglie’s equation,
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
λ = wavelength
Substituting the values in the expression of velocity (v),
= 4.94 × 102 ms–1
v = 494 ms–1
Velocity associated with the neutron = 494 ms–1
Question 2.59:
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.
Answer:
According to de Broglie’s equation,
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ:
λ = 332 pm
Wavelength associated with the electron = 332 pm
Question 2.60:
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:
According to de Broglie’s expression,
Substituting the values in the expression,
Question 2.61:
If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.
Answer:
From Heisenberg’s uncertainty principle,
Where,
Δx = uncertainty in position of the electron
Δp = uncertainty in momentum of the electron
Substituting the values in the expression of Δp:
= 2.637 × 10–23 Jsm–1
Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1)
Uncertainty in the momentum of the electron = 2.637 × 10–23 kgms–1.
Actual momentum
= 1.055 × 10–24 kgms–1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.
Question 2.62:
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = –1/2
2. n = 3, l = 2, ml= 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = –2 , ms = –1/2
5. n = 3, l = 1, ml = –1 , ms= +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2
Answer:
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).