Question 181:
easyThe wavelength of Balmer series of hydrogen atom appears in
The transitions in the Balmer series end on \( n = 2 \). The wavelengths of these transitions lie in the range of 380 nm to 700 nm, which belongs to the visible region of the electromagnetic spectrum.
Question 182:
easyIn an electron microscope electron is accelerated by \(150\text{ kV}\) then de-Broglie wavelength is \(\lambda\). If voltage is increased by \(300\%\) then the de Broglie wavelength of electron will be
Since \(\lambda \propto \frac{1}{\sqrt{V}}\), increasing potential by \(300%\) means \(V' = 4V\). Thus, \(\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}\).
Question 183:
moderateConsider the following statements:
(a) Nuclear density is directly proportional to cube root of mass number.
(b) Binding energy per nucleon is maximum for elements having mass number \(A > 170\).
(c) When two deuterium nuclei fuse together to form a tritium nucleus we get a proton.
(d) Neutrons and protons are bound in a nucleus by the short range weak nuclear force.
The correct statement(s) is/are
Only statement (c) is correct: \(2\text{H} + 2\text{H} \rightarrow 3\text{H} + 1\text{p}\). Nuclear density is independent of mass number, and binding energy per nucleon peaks around Iron \((A=56)\).
Question 184:
moderateA nucleus of mass number 238 at rest emits an alpha particle with kinetic energy \(5.4\text{ MeV}\). The Q value of reaction is
The relation between Q-value and kinetic energy is \(Q = K_{\alpha} \left(1 + \frac{m_{\alpha}}{M_{\text{daughter}}}\right) = 5.4 \times \left(1 + \frac{4}{234}\right) \approx 5.49\text{ MeV}\).
Question 185:
easyAn electron in a hydrogen atom makes a transition from \(n = n_1\) to \(n = n_2\). The time period of revolution of the electron in the initial state is eight times that in final state. The possible value of \(n_1\) and \(n_2\) are
The orbital period is proportional to \(n^3\). Since \(T_1 = 8 T_2\), we must have \(n_1^3 = 8 n_2^3\), which gives \(n_1 = 2n_2\). Thus, \(n_1 = 4\) and \(n_2 = 2\) is correct.
Question 186:
easyIn the following question, a statement of Assertion (A) is followed by a statement of Reason (R).
Assertion (A): On increasing intensity of light, more number of photoelectrons are emitted.
Reason (R): The number of electrons emitted does not depend on intensity of incident radiations
The number of emitted photoelectrons per second is directly proportional to the intensity of incident light above the threshold frequency. Thus, Assertion is true but Reason is false.
Question 187:
easyThe number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be: ($h = 6.6 \times 10^{-34}$ Js)
(2021)
Energy of one photon is $E = \frac{hc}{\lambda}$. The number of photons emitted per second is $n = \frac{P}{E} = \frac{P \lambda}{hc}$. Substituting values: $n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}$.
Question 188:
easyLight of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
(2020)
Initial frequency is $1.5\nu_0$. When the frequency is halved, the new frequency becomes $0.75\nu_0$. Since the new frequency is less than the threshold frequency $\nu_0$, no photoelectric emission will take place, regardless of the intensity. Thus, the photoelectric current will be zero.
Question 189:
easyWhen the light of frequency $2\nu_0$ (where $\nu_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5\nu_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is:
(2018)
Using Einstein's photoelectric equation, $h\nu = h\nu_0 + \frac{1}{2}mv^2$. For the first case, $h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 \implies \frac{1}{2}mv_1^2 = h\nu_0$. For the second case, $h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 \implies \frac{1}{2}mv_2^2 = 4h\nu_0$. Dividing the two equations gives $\frac{v_1^2}{v_2^2} = \frac{1}{4}$, which means $\frac{v_1}{v_2} = \frac{1}{2}$.
Question 190:
easyThe photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is:
(Given $h = 4.14 \times 10^{-15}$ eV and $c = 3 \times 10^8$ ms$^{-1}$)
(2017-Delhi)
Work function $W = \frac{hc}{\lambda_0} = \frac{1242 eV nm}{325 nm} \approx 3.82 eV$. Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1242 eV nm}{253.6 nm} \approx 4.89 eV$. Maximum kinetic energy $K_{max} = E - W = 4.89 - 3.82 = 1.07 eV$. $K_{max} = 1.07 \times 1.6 \times 10^{-19} J \approx 1.7 \times 10^{-19} J$. Also $K_{max} = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2 \times 1.7 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 0.61 \times 10^6 ms^{-1}$.