Photons of frequencies 2.2×1015Hz and 4.6×1015Hz are incident on a metal surface. The corresponding stopping potentials were found to be 6.6 V and 16.5 V respectively. Given e=1.6×10−19 C, the value of universal Planck's constant is
A
6.6×10−34Js
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B
6.7×10−34Js
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C
6.5×10−34Js
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D
6.8×10−34Js
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Solution
The correct option is A6.6×10−34Js From 1st given condition h(2.2×1015)−W=6.6eV -------------- (I) From 2nd condition, h(4.6×10−15)−W=16.5eV-------------(II) Solving the above two simultaneous eqn. we get, h=4.125×10−15evs Converting to Js, h=4.125×10−15×1.6×10−19Js ⇒h=6.6×10−34Js