The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:
[Given: The threshold frequency of platinum is $1 . 3 \times {10}^{15}{s}^{- 1}$ and $h = 6 . 6 \times {10}^{- 34}Js$ ]

Question

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:

[Given: The threshold frequency of platinum is $1 . 3 \times {10}^{15}{s}^{- 1}$ and $h = 6 . 6 \times {10}^{- 34}Js$ ]

Quantrex Academy · Accepted Answer

Correct answer: A. 8 . 58 × 10 ^ - 19 J

Radiation energy = threshold energy of metal + kinetic energy of ejected electron

For minimum energy,

Radiation energy = threshold energy of metal

$E = hνE = 6 . 6 \times {10}^{- 34} \times 1 . 3 \times {10}^{15} = 8 . 58 \times {10}^{- 19}J$

The minimum energy that must be possessed by photons in order to...