The magnetic field associated with a light wave is given, at the origin, by $B = B_{0} [\sin (3.14 \times 10^{7}) c t + \sin (6.28 \times 10^{7}) c t] .$ 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?

$(c = 3 \times 10^{8} m s^{- 1}, h = 6 .6 \times 10^{- 34} J s)$

Question

The magnetic field associated with a light wave is given, at the origin, by

B = B_{0} [\sin (3.14 \times 10^{7}) c t + \sin (6.28 \times 10^{7}) c t] .

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 s)

Aicharya · Accepted Answer

B = B 0 sin &#8289; 3.14 &#215; 10 7 c t + sin &#8289; 6.28 &#215; 10 7 c t Maximum K E will associate with the maximum frequency f max &#8289; &#160; = 6.28 &#215; 10 7 c 2 &#960; = 10 7 &#215; 3 &#215; 10 8 &#160; Hz &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; = 3 &#215; 10 15 &#160; Hz K E m a x = h f m a x - &#981; = 6 .63 &#215; 10 - 34 &#215; 3 &#215; 10 15 1 .6 &#215; 10 - 19 - 4 .7 &#160; eV = 7 .72 &#160; eV

The magnetic field associated with a light wave is given, at the origin, by B = B 0 sin ⁡ 3.14 × 10 7 c t +…

Solution