Question

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Calculate the kinetic energy of the electron ejected when yellow light of frequency 5.2×10¹⁴sec-¹ falls on the surface of potassium metal. Threshold frequency of potassium is 5×10¹sec-¹ .

Answer 1

Frequency of yellow light, fy = 5.2×10^14 Hz

Threshold frequency of Potassium, f0 = 5×10^14 Hz
( I think you have mentioned it as 5×10 by mistake)

Since the frequency (and hence energy) of the incident light is more than the threshold frequency, electron will be emitted from its surface.

The amount of energy which will be used to remove the electron is called work function, given by (h×f0)
The rest amount of energy of the incident light will be used to provide kinetic energy to the emitted electron.

So KE of electron = energy of incident light - work function of potassium

Or KE = h(fy) - h(f0)

Or KE = h×( ft - f0)

Or KE = 6.63×10^(-34) × (5.2×10^14 - 5.0×10^14)

Or KE = 6.63×10^(-20) × 0.2

Or KE = 1.326 × 10^(-20) J


KE of emitted electron is 1.326 × 10^(-20) J

Answer 2

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ĄNsWeR࿐}}}$

Frequency of yellow light, fy = 5.2×10^14 Hz

Threshold frequency of Potassium, f0 = 5×10^14 Hz

( I think you have mentioned it as 5×10 by mistake)

Since the frequency (and hence energy) of the incident light is more than the threshold frequency, electron will be emitted from its surface.

The amount of energy which will be used to remove the electron is called work function, given by (h×f0)

The rest amount of energy of the incident light will be used to provide kinetic energy to the emitted electron.

So KE of electron = energy of incident light - work function of potassium

Or KE = h(fy) - h(f0)

Or KE = h×( ft - f0)

Or KE = 6.63×10^(-34) × (5.2×10^14 - 5.0×10^14)

Or KE = 6.63×10^(-20) × 0.2

Or KE = 1.326 × 10^(-20) J

KE of emitted electron is 1.326 × 10^(-20) J

HOPE SO IT WILL HELP.....