Modern Physics - NEET Physics Questions
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Modern Physics

Question 181: easy

The wavelength of Balmer series of hydrogen atom appears in

1. Infrared region
2. Ultraviolet region
3. Visible region
4. Microwave region
View Answer

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: easy

In 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

1. \(\lambda /2\)
2. \(\lambda\)
3. 2\(\lambda\)
4. \(\lambda /4\)
View Answer

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: moderate

Consider 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

1. Both (a) and (b)
2. Only (b)
3. Only (c)
4. Both (c) and (d)
View Answer

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: moderate

A nucleus of mass number 238 at rest emits an alpha particle with kinetic energy \(5.4\text{ MeV}\). The Q value of reaction is

1. 5.49 MeV
2. 5.88 MeV
3. 5.67 MeV
4. 5.60 MeV
View Answer

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: easy

An 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

1. n_1 = 4, n_2 = 2
2. n_1 = 8, n_2 = 2
3. n_1 = 8, n_2 = 1
4. n_1 = 6, n_2 = 2
View Answer

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: easy

In 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


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true statement but (R) is false
4. Both (A) & (R) are false statements
View Answer

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.