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When a photon of frequency 1.0×10^15s-1 was allowed to hit a metal surface,an electron having 1.988×10^-19j of kinetic energy was emitted.calculate the threshold frequency of the metal.show that an electron will not be emitted if photon of wavelength 600nm hits the metal surface

When a photon of frequency 1.0×10^15s-1 was allowed to hit a metal surface,an electron having 1.988×10^-19j of kinetic energy was emitted.calculate the threshold frequency of the metal.show that an electron will not be emitted if  photon of wavelength 600nm hits the metal surface
 

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
From the Photoelectric effect eqn. => 
hν=hν0 + K. E 
Where hv is the vitality of the photons episode on the metal, hν0 is the edge vitality and K.E is the active vitality with which electron is launched out. 
K.E = 1.988 × 10-19 J 
v = 1 × 1015 s-1 
h (boards steady) = 6.626 × 10-34 Js 
ν0 = hν − K.EhSubstitute the incentive in above equation, we get : ν0 = 6.626 ×10−34 ×1×1015 − 1.988 × 10 −19 6.626 ×10−34 ν0 = 0.700 × 1015 s−1 
Consequently the limit frequeny of electron is 7.00 × 1014 s-1. 
In the event that the esteem that you have given is inaccurate as it is said as 600 nm. Here you should take note of that freqency has a unit of every second except not meter or nm. 
If it's not too much trouble recheck the inquiry, Here for this situation if the we get limit recurrence less that ascertained above then there would be no launch of electron. 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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