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When a metallic surface is illuminated with radiation of wavelength lambda the stopping potential is v. If the same surface is illuminated with rafiation of wavelength 2lambda the stopping potential is v/4. The threshold wavelength for metallic surface? A. 3 lambdaB. 4lambdaC. 5 lambdaPlease give a detailed explanation.

kanika , 7 Years ago
Grade 12th pass
anser 2 Answers
Adarsh

Last Activity: 7 Years ago

let threshold wavelength be λo.  Ans work function be Φ.
so, eV0 = hc/λ – Φ            ….....  (1)
eV0/4 = hc/2λ – Φ             …........ (2) 
on dividing we get , hc/λ – Φ = 2hc/λ – 4Φ 
=> 3Φ=hc/λ
=> Φ = hc/3λ …       so λ0=3λ.
Hence 3λ is the threshold wavelenght ..
 

Yash Chourasiya

Last Activity: 4 Years ago

Dear Student

Case(i),eV = hc/λ − hc/λ0..........(i)
Case (ii),eV4 = hc/2λ − hc/λ0
or eV = 4hc/2λ − 4hc/λ0.......(ii)

From equation (i) & (ii), we get
λ0= 3λ

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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