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The work function of a metal is 4.2 eV ,...

The work function of a metal is `4.2 eV` , its threshold wavelength will be

A

`4000 Å`

B

`3500 Å`

C

`2955 Å`

D

`2500 Å`

Text Solution

Verified by Experts

The correct Answer is:
C
`f_0=phi_0/h=(4.2 xx1.6 xx10^(-19))/(6.6 xx10^(-34))`
`implies lamda_0=c/(f_(0))=(3xx10^8xx6.6xx10^(-34))/(4.2 xx1.6 xx10^(-19))`
`lamda_0~~2946 xx10^(-10)m`
`lamda_0~~2955 Å`
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