Uncategorized - NEET Physics Chapterwise MCQs & PYQs

NEET Uncategorized MCQs & PYQs

Practice NEET Uncategorized Questions

Question 281:

easy

27. A 5 watt source emits monochromatic light of wavelength $5000 \mathring{A}$ . When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of (2007)

The intensity of light $I$ is inversely proportional to the square of the distance $r$ from a point source, so $I \propto \frac{1}{r^2}$ . When the distance is doubled from $0.5 m$ to $1.0 m$ , the intensity becomes $\frac{1}{4}$ of its initial value. Since the number of photoelectrons liberated is directly proportional to intensity, it will also be reduced by a factor of 4.

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Question 282:

44. Work function of a metal surface is $\phi = 1.5 eV$. If a light of wavelength $5000 \AA$ falls on it then the maximum K.E. of ejected electron will be: (1998)

Energy of incident light is $E = \frac{hc}{\lambda} = \frac{12400}{5000} eV = 2.48 eV$. According to Einstein's photoelectric equation, the maximum kinetic energy is $K_{max} = E - \phi = 2.48 eV - 1.5 eV = 0.98 eV$.

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Question 283:

easy

45. Which of the following statement is correct? (1997)

The photoelectric current is directly proportional to the intensity of incident light, provided the frequency of incident light is greater than the threshold frequency.

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Question 284:

easy

51. The cathode of a photoelectric cell is changed such that the work function changes from $W_1$ to $W_2$ ($W_2 > W_1$). If the current before and after changes are $I_1$ and $I_2$, all other conditions remaining unchanged, then (assuming $h\nu > W_2$) (1992)

Saturation photoelectric current depends only on the intensity of the incident light (number of photons per second) and is independent of the work function of the cathode material, provided the incident frequency is above the threshold. Therefore, $I_1 = I_2$.

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Question 285:

easy

84. In a discharge tube at $0.02 mm$, there is formation of (1996)

At a very low pressure of about $0.02 mm$ of Hg in a discharge tube, the Crookes dark space expands to fill the entire tube.

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Question 286:

easy

20. Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model? (2011 Mains)

Energy levels of H-atom are $-13.6 eV$, $-3.4 eV$, $-1.51 eV$, $-0.85 eV$, etc. Possible photon energies are differences between these: $E_3-E_2 = 1.89 \approx 1.9 eV$, $E_4-E_3 = 0.66 \approx 0.65 eV$, $E_{\infty}-E_1 = 13.6 eV$. There is no transition corresponding to $11.1 eV$.

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Question 287:

easy

24. The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between: (2009)

Number of spectral lines emitted is $\frac{n(n-1)}{2} = 6 \Rightarrow n = 4$. Maximum wavelength corresponds to the minimum energy difference. For transitions among levels up to $n=4$, the transition $4 \rightarrow 3$ has the minimum energy difference and thus the maximum wavelength.

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Question 288:

19. The ratio of the radii of the nuclei $ _{13}Al^{27} $ and $ _{52}Te^{125} $ approximately (1990)

The radius of a nucleus is given by $ R = R_0 A^{1/3} $. The ratio of their radii is $ R_{Al} / R_{Te} = (A_{Al} / A_{Te})^{1/3} = (27 / 125)^{1/3} = 3 / 5 = 6 / 10 $.

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Question 289:

43. Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, number of P species are $4 N_0$ and that of Q are $N_0$. Half-life of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be: (2011 Mains)

Equating remaining nuclei: $4N_0(1/2)^{t/1} = N_0(1/2)^{t/2}$, solving gives $2^{t/2} = 4$ so $t = 4$ min. Remaining $P = 4N_0(1/16) = N_0/4$ and $Q = N_0(1/4) = N_0/4$. Formed $R = 5N_0 - (N_0/4 + N_0/4) = 4.5N_0 = \frac{9N_0}{2}$.

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Question 290:

44. The half life of a radioactive isotope ‘X’ is 50 years. It decay to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio 1 : 15 in a sample of a given rock. The age of the rock was estimated to be: (2011 Pre)

Ratio $X/Y = 1/15$, so fraction of X remaining is $X/(X+Y) = 1/16 = (1/2)^4$. This means 4 half-lives have elapsed. The age is $4 \times 50 = 200$ years.

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