Question 171:
easyAn electron jumps from orbit \(n = 4\) to \(n = 3\) in hydrogen atom. Wavelength of the emitted radiation is (\(R\) is Rydberg’s constant)
Using Rydberg's formula, \(\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\). Substituting \(n_1 = 3\) and \(n_2 = 4\) gives \(\frac{1}{\lambda} = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}\). Hence, \(\lambda = \frac{144}{7R}\).
Question 172:
easyA nucleus of nuclear density \(\rho\) disintegrates into two daughter nuclei with masses in the ratio 9:21. Density of the smaller nucleus is
Nuclear density is constant and independent of mass number \(A\) because mass is proportional to \(A\) and volume is also proportional to \(A\). Thus, the density remains \(\rho\).
Question 173:
easyThe ionisation potential of hydrogen is 13.6 V. The energy required to remove an electron from the third orbit of hydrogen is
The energy of an electron in the \(n\)-th orbit of hydrogen is given by \(E_n = -frac{13.6}{n^2}\text{ eV}\). For \(n=3\), \(E_3 = -frac{13.6}{9} = -1.51\text{ eV}\). Thus, the energy required to remove it is 1.51 eV.
Question 174:
easyGiven below are two statements:
Statement-I: The de Broglie wavelength (\(\lambda\)) associated with a moving particle is related to its momentum (\(p\)) as \(\lambda = \frac{h}{p}\).
Statement-II: Photons are electrically neutral and are not deflected by electric and magnetic fields.
In the light of above statements choose the most appropriate answer from the options given below.
Statement-I is the basic de Broglie relationship \(\lambda = h/p\) which is correct. Statement-II is also correct as photons carry no charge and do not interact with electromagnetic fields.
Question 175:
easyIn a hypothetical situation, all the atoms in a hydrogen sample are excited to same state. During de-excitation, photon with lowest energy was found to have \(0.66\text{ eV}\). The photon with the highest energy will have energy equal to
For hydrogen atom, \(E_n - E_{n-1} = 0.66\text{ eV}\) corresponds to \(n = 5\) to \(n = 4\) transition (since \(E_5 - E_4 = -0.85 - (-1.51) = 0.66\text{ eV}\)). The highest energy photon is emitted for transition from \(n = 5\) to \(n = 1\), which is \(E_5 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV}\).
Question 176:
moderateThe de Broglie wavelengths of a proton and an \(\alpha\)-particle are \(3\lambda\) and \(\lambda\) respectively. The ratio of the velocities of proton to \(\alpha\)-particle is
Using the de Broglie wavelength formula \(\lambda = \frac{h}{mv}\), the velocity is (v = \frac{h}{m\lambda}\). Therefore, \(\frac{v_p}{v_\alpha} = \frac{m_\alpha \lambda_\alpha}{m_p \lambda_p} = \frac{4 m_p \cdot \lambda}{m_p \cdot 3\lambda} = \frac{4}{3}\).
Question 177:
easyConsider the following statements:
a. The density of the nuclear matter is independent of the size of the nucleus.
b. Fusion of light nuclei to form heavier nuclei is possible at room temperature.
c. Electron and positron are particle-antiparticle pair.
The correct statement(s) is/are
Nuclear density is constant for all nuclei, hence independent of size (Statement a is true). Nuclear fusion requires extremely high temperatures (Statement b is false). Positron is the antiparticle of electron (Statement c is true).
Question 178:
easyGiven below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Binding energy per nucleon for atoms having mass number between \(30\) to \(170\), almost remains same.
Reason (R): Nuclear force is a short range force and hence get saturated.
In the light of the above statements, choose the correct answer from the options given below.
The binding energy per nucleon is nearly constant in the range \(30 < A < 170\) due to the short-range nature and saturation property of nuclear forces. Thus, both Assertion and Reason are true and the Reason is the correct explanation.
Question 179:
easyThe correct statement regarding nuclear force is
Nuclear forces are very short-range forces (acting within femtometers) and they are charge independent, meaning the force between proton-proton, neutron-neutron, and proton-neutron is approximately the same.
Question 180:
easyIn a photoelectric effect, a metal having threshold frequency ( f_0 ) is used. When it is irradiated with photons of frequency ( 2f_0 ), the stopping potential comes out to be ( V_0 ). If the same metal is irradiated with photons of frequency ( 3f_0 ), then the stopping potential will be equal to
From Einstein's photoelectric equation, \( eV_0 = h(2f_0) - hf_0 = hf_0 \). For frequency ( 3f_0 ), \( eV' = h(3f_0) - hf_0 = 2hf_0 = 2(eV_0) \), which gives \( V' = 2V_0 \).