1.

The work functions for metals A,B and Care 1.92 eV, 2.0 eV and 5.0 eV respctively. The metals which will emit photoelectrons for a radiation of wavelength 4100Å is // are

Answer»

A only
both A and B
all these METALS
none

Solution :Energy of radiation `E = h UPSILON = (hc)/(LAMBDA) = ((6.6 xx 10^(-34) xx 3 xx 10^(8)//1.6 xx 10^(-19))/(410 xx 10^(-9)))=(1240)/(410)`
E = 3.04 eV
Since energy of incident radiation is greater than the work function of metals A and B. So metal A and B will emit photoelectrons.


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