Question 161:
easyAssertion (A): Beryllium can be used as a moderator in nuclear fission reactor.
Reason (R): A fast moving electron on collision with a light stationary particle loses most of its energy in nuclear reactor.
Assertion (A) is true; Beryllium is a suitable moderator due to low mass and absorption.
Reason (R) is false; moderation involves neutrons, not electrons. Electrons are not used for moderating neutrons in nuclear reactors. Thus, (A) is true, but (R) is false.
Question 162:
easyAssertion (A): Nuclear force is short range while gravitation and electric force are universal.
Reason (R): Nuclear force does not follow inverse square law.
Assertion (A) is true; nuclear force is short-range, while gravitational and electric forces are long-range.
Reason (R) is true; nuclear force has a complex dependence, not \(1/r^2\). (R) correctly explains (A) as the absence of inverse square law behavior is characteristic of short-range forces.
Question 163:
easyAssertion (A): In solid each electron will have a different energy level.
Reason (R): In solid crystal each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges.
Due to the Pauli exclusion principle, no two electrons can occupy the same quantum state. In a solid, each electron experiences a unique electrostatic environment. Thus, Assertion (A) is true, and Reason (R) provides the correct explanation for it.
Question 164:
easyAssertion (A): As we increase applied voltage on LED intensity of emitted light first increases then decreases.
Reason (R): We use LED in forward bias.
Assertion (A) is true in a nuanced sense; while intensity generally increases with voltage/current, at very high currents, efficiency can decrease due to thermal effects or Auger recombination (LED droop), leading to a peak and subsequent decrease in intensity. Reason (R) is true; LEDs are operated in forward bias. However, (R) does not explain the intensity variation described in (A).
Question 165:
easyThe ground state energy of hydrogen atom is \(-13.6 \text{eV}\). The energy needed to ionize hydrogen atom from its second excited state will be
Second excited state corresponds to \(n = 3\). The energy of this state is \(E_3 = -\frac{13.6}{3^2} = -1.51 \text{eV}\). Therefore, the energy required to ionize the electron from this level is \(+1.51 \text{eV}\).
Question 166:
easyThe maximum kinetic energy of the emitted photoelectrons in photoelectric effects is independent of:
The maximum kinetic energy of photoelectrons is given by Einstein's equation: \(K_{\text{max}} = h\nu - \Phi\). It depends on frequency, wavelength, and work function, but is independent of the light's intensity, which only affects the number of emitted photoelectrons.
Question 167:
easyThe de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:
The de Broglie wavelength of an electron is \(\lambda = \frac{1.227}{\sqrt{V}} \text{nm}\). Substituting \(V = 81 \text{V}\), we get \(\lambda = \frac{1.227}{9} \approx 0.136 \text{nm}\).
Question 168:
easyThe wavelength of Lyman series of hydrogen atom appears in
Lyman series corresponds to transitions to the ground state (\(n=1\)). These high-energy transitions emit radiation in the ultraviolet region of the spectrum.
Question 169:
moderateGiven below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): \(\text{Fe}^{56}\) nucleus is more stable than \(\text{U}^{235}\) nucleus.
Reason (R): Binding energy of \(\text{Fe}^{56}\) nucleus is more than the binding energy of \(\text{U}^{235}\).
In the light of the above statements, choose the correct answer from the option given below.
\(\text{Fe}^{56}\) is more stable because its binding energy per nucleon is higher than that of \(\text{U}^{235}\). However, the total binding energy of \(\text{U}^{235}\) is greater than that of \(\text{Fe}^{56}\) because it contains many more nucleons. Hence, Assertion is true but Reason is false.
Question 170:
easyUltraviolet radiation of \(6.1 \text{eV}\) falls on an aluminium surface of work function \(4.2 \text{eV}\). The kinetic energy of the fastest electron emitted is
According to Einstein's photoelectric equation, \(K_{max} = h\nu - \phi = 6.1 \text{eV} - 4.2 \text{eV} = 1.9 \text{eV}\). Converting to Joules: \(1.9 \times 1.6 \times 10^{-19} \text{J} = 3.04 \times 10^{-19} \text{J}\).