QJEEM19Ph0f13: JEE Main 2019 Physics - Class 12 JEEM19Ph0f13 Solution: The magnetic field associated with a light wave is given, at…

Given that,

B = B₀[sin (3.14 $\times$ 10⁷) ct + sin(6.28 $\times$ 10⁷) ct]

This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ( $\phi$ ) = 4.7 eV

Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As,

K_max = E_max $-$ $\phi$

Emax = hF

= h $\times$ ${\omega \over {2\pi }}$

= 6.6 $\times$ 10 ^{$-$ 34} $\times$ ${{6.28 \times {{10}^7} \times 3 \times {{10}^8}} \over {2\pi }}$

= 6.6 $\times$ 3 $\times$ 10 ^{$-$ 19} J

= ${{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV$

= 12.375 $eV$

$\therefore$   K_max = 12.375 $-$ 4.7

= 7.675 $eV$

$\simeq$   7.7 $eV$