Dual Nature of Radiation and Matter
- Gases begin to conduct electricity electricity at low pressure because
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The ionisation requires high energy electrons.
Correct Option: B
The ionisation requires high energy electrons.
- An ionization chamber with parallel conducting plates as anode and cathode has 5 × 107 electrons and the same number of singly charged positive ions per cm3. The electrons are moving towards the anode with velocity 0.4 m/s. The current density from anode to cathode is 4μA/m2. The velocity of positive ions moving towards cathode is
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Current = Ie + Ip
Ie and Ip are current due to electrons and positively charged ions.
I = neAVd
[ n = 5 × 107 /cm3 = 5 × 107 × 106 / m3 = 5 × 1013/m3 ]
Ie = 5 × 1013 × 1.6 × 10-19 × A × 0.4
Ip = 5 × 1013 × 1.6 × 10-19 × A × v
I = Ie + Ip
= 5 × 1013 × 1.6 × 10-19 × A (v + 0.4)
Given,
I/A = 4 × 10-6 A/m2
4 × 10-6 A =5 × 10-6 × 1.6 × A (v + 0.4)4 = v + 0.4 8
⇒ 0.5 = v + 0.4 ⇒ v = 0.1 m/sCorrect Option: D
Current = Ie + Ip
Ie and Ip are current due to electrons and positively charged ions.
I = neAVd
[ n = 5 × 107 /cm3 = 5 × 107 × 106 / m3 = 5 × 1013/m3 ]
Ie = 5 × 1013 × 1.6 × 10-19 × A × 0.4
Ip = 5 × 1013 × 1.6 × 10-19 × A × v
I = Ie + Ip
= 5 × 1013 × 1.6 × 10-19 × A (v + 0.4)
Given,
I/A = 4 × 10-6 A/m2
4 × 10-6 A =5 × 10-6 × 1.6 × A (v + 0.4)4 = v + 0.4 8
⇒ 0.5 = v + 0.4 ⇒ v = 0.1 m/s
- The photoelectric threshold wavelength of silver is 3250 × 10–10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is
(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)
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Both answers are correct Given,
λ0 = 3250 × 10–10 m
λ = 2536 × 10–10 m φ = hc = 4.14 × 10-15 × 3 × 108 = 3.82eV λ0 3250 × 10-10 hv = hc = 4.14 × 10-15 × 3 × 108 = 4.89eV λ 2536 × 10-10
According to Einstein's photoelectric equation,
Kmax = hv – φ
KEmax = (4.89–3.82)eV=1.077 eV1 = 1.077 × 1.6 × 10–19 2 ⇒ v = 
2 × 1.077 × 1.6 × 10-19 9.1 × 10-31
or, v = 0.6 × 106 m/s or 6 × 105 m/sCorrect Option: E
Both answers are correct Given,
λ0 = 3250 × 10–10 m
λ = 2536 × 10–10 m φ = hc = 4.14 × 10-15 × 3 × 108 = 3.82eV λ0 3250 × 10-10 hv = hc = 4.14 × 10-15 × 3 × 108 = 4.89eV λ 2536 × 10-10
According to Einstein's photoelectric equation,
Kmax = hv – φ
KEmax = (4.89–3.82)eV=1.077 eV1 = 1.077 × 1.6 × 10–19 2 ⇒ v = 2 × 1.077 × 1.6 × 10-19 9.1 × 10-31
or, v = 0.6 × 106 m/s or 6 × 105 m/s
- Momentum of a photon of wavelength λ is
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According to de-Brogie wave equation,
λ = h = h mv P ∴ P = h λ Correct Option: A
According to de-Brogie wave equation,
λ = h = h mv P ∴ P = h λ
- The X-rays cannot be diffracted by means of an ordinary grating because of
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We know that the X-rays are of short wavelength as compared to grating constant of optical grating. As a result of this, it makes difficult to observe X-rays diffraction with ordinary grating.
Correct Option: B
We know that the X-rays are of short wavelength as compared to grating constant of optical grating. As a result of this, it makes difficult to observe X-rays diffraction with ordinary grating.
