Question

Hey guys

Answer this plz :)

Don't spam

When light with a frequency 547.5 THz illuminates a metallic surface, the most energetic photoelectrons have 1.260 x 10^-19 J of kinetic energy. When light with a frequency of 738.8 THz is used instead, the most energetic photoelectrons have 2.480 x 10^-19 J of kinetic energy. Using these experimental results. determine the approximate value of Planck's constant.​

Answer 1

hope this helps you Vishal

Answer 2

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$The \:  maximum \: kinetic \: energy \:  of \: the \: photoelectrons \: is \: given \: by \: the \: formula \:  K_M \\ =hf−ϕ . \\ We have two situations where for  \\f_1=547.5\times10^{12}Hzf1 \\ =547.5×1012Hz we get K_{M1} \\ =1.26\times10^{-19}JKM1 \\ =1.26×10−19J and \\ for f_2=738.8\times10^{12}Hzf2 \\ =738.8×1012Hz we get K_{M2} \\ =2.48\times10^{-19}JKM2 \\ =2.48×10−19J , so we have: \\ K_{M1}=hf1−ϕ\\ K_{M2}M2=hf2−ϕWe \: can \: eliminate \:  \phiϕ  \: by \: substracting \: the \: first \: equation \: to \: the \: second: \\ K_{M2}-K_{M1} \\ =hf_2-\phi-(hf_1-\phi) \\ =h(f_2-f_1)KM2−KM1 \\ =hf2−ϕ−(hf1−ϕ)=h(f2−f1) \\ Which means: \\ h=\frac{K_{M2}-K_{M1}}{f_2-f_1} \\ =\frac{2.48\times10^{-19}J-1.26\times10^{-19}J}{738.8\times10^{12} -547.5\times10^{12}Hz} \\ =6.377\times10^{-34}kgm^2/sh=f2−f1KM2−KM1 \\=738.8×1012Hz−547.5×1012Hz2.48×10−19J−1.26×10−19J \\ =6.377×10−34kgm2/s$

$\boxed{I \:Hope\: it's \:Helpful}$

${\sf{\bf{\blue{@ℐᴛz ᴅɪɴᴜ࿐}}}}$