If light of wavelength `412.5` nm is incident on each of the metals given below, which one will show photoelectric emission and why ?

Explanation

Calculating the energy of the
incident phohton
Identifying the metals
Reason
The energy of a photon of incident radiation is given by
` E= ( h_in ) /( lambda ) `
` therefore E= ( 6.63xx 10 ^( -34) xx 3xx10 ^(8) ) /( ( 412.5xx 10 ^(-9)) xx (1.6xx10 ^(-19)) eV `
`~= 3.01eV `
Hence ,only Na and K will show photoelectric emission
[Note : Award this mark even if the student wirtes the name of only one of these metals ] Reason : They energy of the incident photon is more that the work function of only these two metals.