Q4. Assertion: for light of one frequency, the stopping potential is the same at every intensity. Reason: the number of photoelectrons emitted per second is proportional to the intensity. Are both true, and does the reason explain the assertion?
Both are true, but the reason does not explain the assertion. The reason explains why the saturation current grows with intensity; the stopping potential depends on the maximum kinetic energy, which intensity does not change (NCERT pages 278–279).
Q5. In one experiment the saturation current is 12 μA. How many photoelectrons reach the collector each second?
n = I/e = (12 × 10−6)/(1.6 × 10−19) = 7.5 × 1013 per second (reasoned from NCERT page 278).
Q6. The fastest photoelectrons have a speed of 6.0 × 105 m/s. Find the stopping potential.
V0 = mv2/(2e) = (9.11 × 10−31 × 3.6 × 1011)/
Q7. Two metals A and B give parallel lines on a graph of stopping potential against frequency, A’s meeting the axis at a lower frequency. Which metal needs less energy to release an electron?
A. Its threshold frequency is lower, so light of lower frequency already ejects electrons from it. NCERT links the two in Section 11.6: the work function is φ0 = hν0 (NCERT pages 279–280 and 282).