DP Physics (first assessment 2025)

Question 21M.2.HL.TZ1.b

[Maximum mark: 2]

21M.2.HL.TZ1.b

Show that the maximum velocity of the photoelectrons is $700 {km s}^{- 1}$ .

[2]

Markscheme

« $e V = \frac{1}{2} m v^{2}$ » and manipulation to get $v$ ✓

$v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 1.4}{9.1 \times 10^{- 31}}}$ OR $702 « {km s}^{- 1} »$ ✓

Must see either complete substitution or calculation to at least 3 s.f. for MP2

Syllabus sections