NEET Practice Questions (MCQs) with Answers & Solutions

The text states: 'For example, in the familiar phenomenon of seeing an object by our eye, both descriptions are important. The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture.'

According to the provided text, 'For a frequency $\nu$ of incident radiation, lower than the cut-off frequency $\nu_0$, no photoelectric emission is possible even if the intensity is large.' This directly contradicts option 4.

The text states: 'For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light (Fig. 11.2)'. This means it increases linearly.

The context mentions, 'the maximum kinetic energy of photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity.' Also, from Fig 11.3 description, 'the stopping potential remains the same as that for the incident radiation of intensity $I_1$ ... Thus, for a given frequency of the incident radiation, the stopping potential is independent of its intensity.' Since $K_{max} = eV_0$, $K_{max}$ is also independent of intensity.

The text defines it as: 'This minimum, cut-off frequency $\nu_0$, is called the threshold frequency. It is different for different metals.' And 'Below a certain frequency (threshold frequency) $\nu_0$, characteristic of the metal, no photoelectric emission takes place, no matter how large the intensity may be.'

The text explicitly states: 'It is now known that emission starts in a time of the order of $10^{-9}$ s or less.' This indicates the instantaneous nature of the process.

The text mentions: 'However, some alkali metals such as lithium, sodium, potassium, caesium and rubidium were sensitive even to visible light.' Zinc and cadmium are mentioned as responding to ultraviolet light.

Einstein’s photoelectric equation is given as: '$1/2 m v^2_{max} = V_0 e = h\nu – \phi_0 = h (\nu – \nu_0)$'. Here, $\phi_0$ represents the work function, which is $h\nu_0$.

Point 4 in the summary states: 'The stopping potential ($V_0$) depends on (i) the frequency of incident light, and (ii) the nature of the emitter material. For a given frequency of incident light, it is independent of its intensity.'

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.