Chemistry MCQs for NEET — Practice Questions with Answers

Figure 11.4 and its description state: 'The stopping potential is more negative for higher frequencies of incident radiation.' This implies that as frequency increases, the stopping potential increases in magnitude (becomes more negative). This is also consistent with $V_0 = (h/e)(\nu - \nu_0)$, where $h/e$ is positive.

Point 6 in the summary states: 'The classical wave theory could not explain the main features of photoelectric effect. Its picture of continuous absorption of energy from radiation could not explain the independence of $K_{max}$ on intensity, the existence of $\nu_0$ and the instantaneous nature of the process.' All listed options are correct reasons for its failure.

The text describes the setup: 'The polarity of the plates C and A can be reversed by a commutator. Thus, the plate A can be maintained at a desired positive or negative potential with respect to emitter C.'

Figure 11.3 demonstrates this: 'We note that the saturation currents are now found to be at higher values' when intensity is increased (I3 > I2 > I1 corresponds to higher saturation currents). The photoelectric current is directly proportional to the number of photoelectrons emitted per second, which in turn is proportional to the intensity.

The text clearly states: 'Photoelectric effect involves conversion of light energy into electrical energy.'

The context states: 'ultraviolet light gives rise to photoelectric effect in copper while green or red light does not.' This indicates that the frequency of ultraviolet light is above copper's threshold frequency, while the frequencies of green and red light are below it, as lower frequencies mean longer wavelengths and vice versa.

If $\nu < \nu_0$, then $h\nu < h\nu_0$, which means $h\nu - \phi_0$ would be negative. Since kinetic energy cannot be negative, this indicates that no photoelectric emission takes place, consistent with the definition of threshold frequency.

As per the NCERT text, 'In general, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d orbital each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series.' This means atomic radii generally decrease across a period due to increasing effective nuclear charge, as the shielding effect of d electrons is not very effective.

The NCERT states, 'The radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction'. This contraction essentially compensates for the expected increase in atomic size with increasing atomic number.

The NCERT text explicitly mentions, 'The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their occurrence together in nature and for the difficulty faced in their separation.'