Answer:
(i) The slope \[{{V}_{0}}-v\] graph gives the value of \[\frac{h}{e},\] which is same for both materials \[{{M}_{1}}\]and \[{{M}_{2}}\]. (ii) According to Einstein's photoelectric equation, \[{{K}_{\max }}=hv-{{W}_{0}}=hv-h{{v}_{0}}\] The material A^ has a lower value of threshold frequency\[{{v}_{0}}\]. So \[{{M}_{1}}\] will emit photoelectrons of greater kinetic energy for the same frequency \[v\] of the incident radiation.
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