JEE PYQ: Experimental Physics Question 22
Question 22 - 2019 (09 Jan 2019 Shift 1)
The magnetic field associated with a light wave is given at the origin by $B = B_0 [\sin(3.14 \times 10^7)ct + \sin(6.28 \times 10^7)ct]$. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photoelectrons? ($c = 3 \times 10^8$ m s$^{-1}$, $h = 6.6 \times 10^{-34}$ J$\cdot$s)
(1) 6.82 eV
(2) 12.5 eV
(3) 8.52 eV
(4) 7.72 eV
Show Answer
Answer: (4)
Solution
Two EM waves: $\nu_1 = \frac{10^7}{2} \times c$ and $\nu_2 = 10^7 \times c$. KE is maximum for higher frequency photon. $E_{ph} = h\nu_2 = 6.6 \times 10^{-34} \times 10^7 \times 3 \times 10^8 = 12.375$ eV. $KE_{\max} = 12.375 - 4.7 \approx 7.7$ eV.
BETA
