JEE physics

The work function of sodium is 2.3 eV — JEE Physics

TSTushar Saxena · 10 Asked 2mo ago 289 views 1 answer

The work function of sodium is 2.3 eV. Find the maximum kinetic energy of photoelectrons when light of frequency $7.5 \times 10^{14}$ Hz falls on it. ($h = 6.63 \times 10^{-34}$ J·s)

1 Answer

NRNikhil Rao ✓ Accepted · 2mo ago ▲ 31

$$E_{photon} = hf = 6.63\times10^{-34} \times 7.5\times10^{14} = 4.97\times10^{-19} J = 3.11 eV$$

$$KE_{max} = E_{photon} - \phi = 3.11 - 2.3 = 0.81 eV$$

Log in to post your own answer or join the discussion.

Discussion (2)

N
Tip for others: you can verify the answer quickly by checking the units.
NabinThapa70 · 2mo ago
C
Does this approach generalise to the JEE Advanced version of this question?
ChloeLefevre13 · 2mo ago
← Back to all questions