Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Nucleus having more binding energy per nucleon is more stable.
Reason (R): Stability increase with increase in number of nucleons.
In the light of the above statements, choose the correct answer from the options given below.
1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. (A) is false but (R) is true
View Answer
Assertion is true because binding energy per nucleon is the direct measure of nuclear stability. Reason is false because stability does not simply increase with nucleon number; heavy nuclei with a very large number of nucleons become unstable.
The number of photons per second on an average emitted by the source of monochromatic light of wavelength \(600\text{ nm}\), when it delivers the power of \(3.3 \times 10^{-3}\text{ watt}\) will be (\(h = 6.6 \times 10^{-34}\text{ J s}\))
1. \(10^{15}\)
2. \(10^{18}\)
3. \(10^{17}\)
4. \(10^{16}\)
View Answer
Power \(P = n \frac{hc}{\lambda}\), where \(n\) is the number of photons emitted per second. Substituting the given values, we get \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).
An electromagnetic wave of wavelength \(lambda\) is incident on a photosensitive surface of negligible work function. If \(m\) is mass of photoelectron emitted from the surface has de-Broglie wavelength \(lambda_d\), then
1. \(\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2\)
2. \(\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2\)
3. \(\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2\)
4. \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\)
View Answer
With a negligible work function, the maximum kinetic energy of the emitted photoelectron is \(E = \frac{hc}{\lambda}\) and its de-Broglie wavelength is \(\lambda_d = \frac{h}{\sqrt{2mE}}\). Substituting \(E\) gives \(\lambda_d^2 = \frac{h\lambda}{2mc}\), which simplifies to \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\).
The potential difference that must be applied to stop the fastest moving photoelectrons emitted by a metal surface, having work function \(4.5\text{ eV}\), when ultraviolet light of \(2000A^\circ\) falls on it, will be
1. -0.7 V
2. -1.7 V
3. -1.2 V
4. -0.8 V
View Answer
Formula: \(eV_0 = E - \Phi_0\), where \(E = \frac{hc}{\lambda} = \frac{12400}{2000} = 6.2\text{ eV}\). Thus, the stopping potential \(V_0 = 6.2 - 4.5 = 1.7\text{ V}\), requiring an applied potential of \(-1.7\text{ V}\).
When \(^{235}_{92}\text{U}\) undergoes fission, 0.2% of its original mass is changed into energy. How much energy is released if 1 kg of \(^{235}_{92}\text{U}\) undergoes fission?
1. \[18 \times 10^{11} J\]
2. \[6 \times 10^{5} J\]
3. \[18 \times 10^{13} J\]
4. \[18 \times 10^{10} J\]
View Answer
Formula: \(E = \Delta m c^2\). The mass defect is \(\Delta m = 0.2\%\ of 1 kg = 0.002\text{ kg}\). Thus, \(E = 0.002 \times (3 \times 10^8)^2 = 18 \times 10^{13}\text{ J}\).
Consider a hypothetical hydrogen like atom. Wavelength of spectral lines corresponding to transition of electron from \( n = m \) to \( n = 1 \) are given by \( \lambda = \frac{1800 m^2}{m^2 – 1} \text{ \A^\circ} \) \( (m = 2, 3, 4, 5, \dots, \infty) \). Minimum energy of emitted photon is nearly
1. 5.2 eV
2. 7.03 eV
3. 10.2 eV
4. 13.6 eV
View Answer
Minimum energy corresponds to the maximum wavelength, which occurs at \( m = 2 \). Thus, \( \lambda_{\max} = \frac{1800 \times 2^2}{2^2 - 1} = 2400 \text{ \A^\circ} \). The minimum energy is \( E_{\min} = \frac{12400}{\lambda_{\max}} = \frac{12400}{2400} \approx 5.17 \text{ eV} \approx 5.2 \text{ eV} \).
1. The conversion of neutron into proton along with neutrino and electron
2. The conversion of proton into neutron along with antineutrino and electron
3. The conversion of neutron into proton along with positron and antineutrino
4. The conversion of proton into neutron along with neutrino and positron
View Answer
In \( \beta^+ \) decay, a proton inside the nucleus converts into a neutron, emitting a positron (\( e^+ \)) and a neutrino (\( \nu_e \)). The nuclear reaction is \( p \rightarrow n + e^+ + \nu_e \).