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Question: Answered & Verified by Expert
A light of wavelength 3000 Ao falls on a metal surface. Ejected e- is further accelerated by a potential difference of 2V , then final K.E of the e- is found to be 8×10-19J . If threshold energy for the metal surface is 'ϕ'eV . Then find the numerical value of 8ϕ
ChemistryStructure of AtomJEE Main
Solution:
2239 Upvotes Verified Answer
The correct answer is: 9.00
KE final = E photon + E potential ϕ
where ϕ = threshold energy for metal surface
E photon = hc λ = 6.6× 10 34 ×3× 10 8 3000× 10 10
6.6× 10 19 J=4.12eV
E potential =2eV
KE final = 8× 10 19 J 1.6× 10 19 J =5eV
Hence, ϕ=(4.12+25)eV=1.125eV
Numerical value = 8×1.125=9

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