4) Light of frequency 6.80 × 10 14 Hz ejects electrons from surface (A) with a m

Question

4) Light of frequency 6.80 × 1014 Hz ejects electrons from surface (A) with a maximum kinetic energy that is 3.60 × 10-19 J greater than the maximum kinetic energy of electrons ejected from surface B. Calculate the difference in work function for these two surfaces.

5. Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.62 eV to be ejected from a metal whose work function is 2.55 eV. What will be the maximum kinetic energy with which this same radiation ejects electrons from another metal whose work function is 2.17 eV?

Explanation / Answer

4)

frequency , f = 6.8 *10^14 Hz

maximum kinetic energy difference = 3.6 *10^-19 J

energy of photon = h * f

energy of photon = 6.8 *10^14 * 6.626 *10^-34

energy of photon = 4.506 *10^-19 J

the difference in work function = energy of photon - difference in kinetic energy

difference in work function = 4.506 *10^-19 - 3.6 *10^-19

difference in work function = 9.06 *10^-20 J

5)

for the first surface

energy of photon = work fucntion + maximum kinetic energy of electron

energy of photon = 0.62 eV + 2.55 eV

energy of photon = 3.17 eV

now , for the another metal

energy of photon = work fucntion + maximum kinetic energy of electron

3.17 = maximum kinetic energy of electron + 2.17

maximum kinetic energy of electron = 1 eV

the maximum kinetic energy of electron is 1. eV

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