Question

The threshold frequency v0 for a metal is 6*10^14s. Calculate the ke of an electron emitted when radiation oof frequency v = 1.1 *10^15s^-1

Answer 1

Answer:

Explanation:

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 //

Hope it helps you....

Answer 2

Explanation:

don't know ...........