More practice questions

  1. Q4. Assertion: the graph of stopping potential against frequency has the same slope for every metal. Reason: the slope equals h/e. Are both true, and does the reason explain the assertion?

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    Both are true, and the reason explains the assertion: h and e are universal constants, so the slope is the same for all metals; only the threshold changes (NCERT Eq. (11.4), page 282).

  2. Q5. Light of frequency 7.21 × 1014 Hz ejects electrons with a maximum speed of 6.0 × 105 m/s. Find the threshold frequency. (NCERT Exercise 11.8)

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    hν0 = hν − ½mv2. ½mv2 = ½ × 9.11 × 10−31 × 3.6 × 1011 ≈ 1.64 × 10−19 J, which is h × 2.47 × 1014 Hz. So ν0 ≈ 7.21 × 1014 − 2.47 × 1014 ≈ 4.74 × 1014 Hz (NCERT Exercise 11.8, page 289). NCERT’s answer on page 351, 4.73 × 1014 Hz, differs only in the last digit.

  3. Q6. An argon laser gives light of 488 nm, and the stopping potential is 0.38 V. Find the work function of the emitter. (NCERT Exercise 11.9)

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    hc/λ = (6.63 × 10−34 × 3 × 108)/(488 × 10−9) ≈ 4.076 × 10−19 J, which is 2.54 eV with 1 eV = 1.602 × 10−19 J, so φ0 ≈ 2.54 − 0.38 = 2.16 eV (NCERT Exercise 11.9, page 289; answer on page 351). With 1 eV = 1.6 × 10−19 J the same working gives 2.17 eV.

  4. Q7. Two metals have threshold frequencies 4 × 1014 Hz and 6 × 1014 Hz. Light of 8 × 1014 Hz falls on both. Find the ratio of their maximum kinetic energies.

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    Kmax = h(ν − ν0): h × 4 × 1014 against h × 2 × 1014, a ratio of 2 : 1 (NCERT Summary point 7, page 287).