Question 141:
easy41. The current conduction in a discharge tube is due to: (1999)
In a gas discharge tube, the applied high voltage ionizes the gas atoms. This creates free electrons and positive ions. Both of these charged particles migrate towards opposite electrodes under the electric field, contributing to the current conduction.
Question 142:
easy42. Light of wavelength $3000 \mathring{A}$ in Photoelectric effect gives electron of max. K.E. 0.5 eV. If wavelength change to $2000 \mathring{A}$ then max. K.E. of emitted electrons will be: (1999)
Incident energy $E = \frac{hc}{\lambda}$ . Initially, $E_1 = \frac{12400}{3000} \approx 4.13 eV$ . Work function is $W = E_1 - K_1 = 4.13 - 0.5 = 3.63 eV$ . For $2000 \mathring{A}$ , new energy is $E_2 = \frac{12400}{2000} = 6.2 eV$ . The new maximum kinetic energy is $K_2 = E_2 - W = 6.2 - 3.63 = 2.57 eV$ , which is clearly greater than 0.5 eV.
Question 143:
easy43. If the light of wavelength $\lambda$ is incident on metal surface, the ejected fastest electron has speed v. If the wavelength is changed to $\frac{3\lambda}{4}$ the speed of the fastest emitted electron will be: (1998)
Initially, $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - W$ . Finally, $\frac{1}{2}mv'^2 = \frac{hc}{3\lambda/4} - W = \frac{4hc}{3\lambda} - W$ . This can be rewritten as $\frac{1}{2}mv'^2 = \frac{4}{3}(\frac{hc}{\lambda} - W) + \frac{W}{3} = \frac{4}{3}(\frac{1}{2}mv^2) + \frac{W}{3}$ . Since $W$ is positive, $\frac{1}{2}mv'^2 > \frac{4}{3}(\frac{1}{2}mv^2)$ , meaning $v'^2 > \frac{4}{3}v^2$ or $v' > \sqrt{\frac{4}{3}}v$ .
Question 144:
easy45. 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.
Question 145:
easy51. 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, $I_1 = I_2$.
Question 146:
easy84. 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.
Question 147:
easy20. 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 $11.1 eV$.
Question 148:
easy24. 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.