Light of wavelength 1824 1824mathop -text-A-limits-text-0- - incident on the surface of a metal produces photoelectrons with maximum energy 5-3 eV- Now- when the light of wavelength 1216mathop -text-A-limits-text-0- is used- the maximum energy of photoelectrons is 8-7 eV- The work of the metallic surface is - 1-5 eV13-6 eV6-8 eV3-5 eV

Question

Light of wavelength 1824 1824mathop -text-A-limits-text-0-  - incident on the surface of a metal produces photoelectrons with maximum energy 5-3 eV- Now- when the light of wavelength 1216mathop -text-A-limits-text-0-  is used- the maximum energy of photoelectrons is 8-7 eV- The work of the metallic surface is - 1-5 eV13-6 eV6-8 eV3-5 eV

Toppr Admin · Accepted Answer

As we know-E-W0-Kmax- From the given data E is 6-75-xA0-eV -for -x3BB-1824oA- or 10-17-xA0-eV -for -x3BB-1216oA-xA0-x2234-W0-E-x2212-Kmax-6-75-x2212-5-3-1-48-xA0-eVORW0-10-17-x2212-8-7-1-5-xA0-eV

Light of wavelength 1824 18240A , incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength 12160A is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is : 1.5 eV13.6 eV6.8 eV3.5 eV

As we know,E=W0+Kmax. From the given data E is 6.75 eV (for λ=1824oA) or 10.17 eV (for λ=1216oA) ∴W0=E−Kmax=6.75−5.3=1.48 eVORW0=10.17−8.7=1.5 eV

As we know,
$E = W_{0} + K_{m a x}$ . From the given data $E$ is $6.75 e V$ (for $λ = 1824 o A$ ) or $10.17 e V$ (for $λ = 1216 o A$ )
$∴ W_{0} = E - K_{m a x} = 6.75 - 5.3 = 1.48 e V$
OR
$W_{0} = 10.17 - 8.7 = 1.5 e V$