Question

if the photo emissive surface is threshold frequency of 4.6 ×10^14Hz.calculate the energy of photon in electron volt .

if you know then you will give ......​

Answer 1

K.E ( kinetic energy ) = h ( f - f₀ ) 

            h ----- > plank's constant

            f₀ ------> Threshold frequency

            f --------> Frequency 

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          h = 6.626 × 10⁻³⁴

          f₀ = 7× 10¹⁴ Hz

          f = 1 × 10¹⁵ Hz

K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵)  - (7× 10¹⁴)]

       = 1.987 × 10⁻¹⁹ J //

Answer 2

ⓗⓔⓨⓐ

E = hVo

= 6.6 × 10^-34 × 4.6× 10^14

= 30.36 × 10^-20 j

= 30.36 × 10^-20/1.6×10^-19

= 1.897eV

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