Modern Physics - NEET Physics Questions
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Modern Physics

Question 51: moderate

Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct :

1. A
2. B
3. C
4. D
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Question 52: easy

Nuclear forces are :

1. Short ranged attractive and charge independent
2. Short ranged attractive and charge dependent
3. Long ranged repulsive and charge independent
4. Long ranged repulsive and charge dependent
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Question 53: difficult

The binding energy of deuteron \[_{1}^{2}H\] is 1.112 MeV per nucleon and an α-particle

\[_{2}He^{4}\] has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction

\[_{1}^{2}H+_{1}^{2}H\longrightarrow _{2}^{4}He+Q\] , the energy Q released is :

1. 1 MeV
2. 11.9 MeV
3. 23.8 MeV
4. 931 MeV
View Answer
Question 54: easy

Which reaction is not the part of proton-proton cycle ?

1. \(_1 \text{H}^1 + _1 \text{H}^1 \rightarrow _1 \text{H}^2 + beta^+ + nu + Q\)
2. \(_1 \text{H}^2 + _1 \text{H}^2 \rightarrow _2\text{He}^3 + _0\text{n}^1 + Q\)
3. \(_1\text{H}^2 + _1\text{H}^1 \rightarrow _2\text{He}^3 + Q\)
4. \(_2 \text{He}^3 + _2 \text{He}^3 \rightarrow _2 \text{He}^4 + 2(_1 \text{H}^1) + Q\)
View Answer

In the stellar proton-proton chain reaction, deuterium fuses with a proton to form Helium-3, but deuterium does not fuse directly with another deuterium nucleus. Thus, reaction (2) is not part of this cycle.

Question 55: easy

The 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

1. \(1.51\text{ eV}\)
2. \(3.4\text{ eV}\)
3. \(13.6\text{ eV}\)
4. \(6.8\text{ eV}\)
View Answer

Second excited state corresponds to \(n = 3\). The energy is \(E_3 = -\frac{13.6}{3^2} = -1.51\text{ eV}\). The ionization energy required is \(E_{\text{ion}} = 0 - E_3 = 1.51\text{ eV}\).

Question 56: easy

The maximum kinetic energy of the emitted photoelectrons in photoelectric effects is independent of:

1. Frequency of incident radiation
2. Wavelength of incident radiation
3. Work function of material
4. Intensity of incident radiation
View Answer

According to Einstein's photoelectric equation, \(K_{\text{max}} = h\nu - \phi\). The maximum kinetic energy depends on the frequency/wavelength of the incident light and the work function, but is independent of the intensity of the light.

Question 57: easy

The wavelength of Lyman series of hydrogen atom appears in

1. Ultraviolet region
2. Infrared region
3. Visible region
4. Far infrared region
View Answer

The Lyman series transitions terminate at the ground state (\(n = 1\)). The photon energies emitted in these transitions correspond to the ultraviolet region of the electromagnetic spectrum.

Question 58: easy

The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

1. 1.36 nm
2. 0.136 nm
3. 13.6 nm
4. 136 nm
View Answer

The de Broglie wavelength for an electron accelerated through a potential \(V\) is given by \(\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}\). Substituting \(V = 81\text{ V}\), we get \(\lambda = \frac{1.227}{9}\text{ nm} \approx 0.136\text{ nm}\).

Question 59: easy

Ratio of shortest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

1. \(\frac{1}{4}\)
2. \(\frac{1}{9}\)
3. \(\frac{5}{27}\)
4. \(\frac{16}{27}\)
View Answer

The shortest wavelength in a series corresponds to transition from \(n_2 = \infty\) to \(n_1\). For Lyman series, \(\lambda_L = \frac{1}{R}\). For Balmer series, \(\lambda_B = \frac{4}{R}\). Therefore, the ratio \(\frac{\lambda_L}{\lambda_B} = \frac{1}{4}\).

Question 60: easy

Photocell is illuminated by a point source of light, which is placed at a distance \(d\) from the cell. If the distance becomes \(2d\), then number of electrons emitted per second will be

1. Remain same
2. Four times
3. Two times
4. One-fourth
View Answer

Intensity of light from a point source is inversely proportional to the square of the distance, \(I \propto \frac{1}{d^2}\). Since the number of photoelectrons emitted per second is proportional to intensity, doubling the distance reduces the emission to one-fourth.