Light of frequency 1.6 times 10^{16} Hz fall | vedclass

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(B) According to Einstein's photoelectric equation,$K.E. = h(v - v_0)$.For frequency $v_1 = 1.6 	imes 10^{16} \ Hz$,$K.E._1 = h(1.6 	imes 10^{16} -

Vedclass · Accepted Answer

$4 	imes 10^{15}$

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(B) According to Einstein's photoelectric equation,$K.E. = h(v - v_0)$.For frequency $v_1 = 1.6 	imes 10^{16} \ Hz$,$K.E._1 = h(1.6 	imes 10^{16} - v_0) \longrightarrow 	ext{Eq. } (i)$.For frequency $v_2 = 1.0 	imes 10^{16} \ Hz$,$K.E._2 = h(1.0 	imes 10^{16} - v_0) \longrightarrow 	ext{Eq. } (ii)$.Given that $K.E._1 = 2 	imes K.E._2$,we have:$h(1.6 	imes 10^{16} - v_0) = 2 	imes h(1.0 	imes 10^{16} - v_0)$.$1.6 	imes 10^{16} - v_0 = 2.0 	imes 10^{16} - 2v_0$.$2v_0 - v_0 = 2.0 	imes 10^{16} - 1.6 	imes 10^{16}$.$v_0 = 0.4 	imes 10^{16} \ Hz = 4 	imes 10^{15} \ Hz$.

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