A) exactly \[\frac{({{E}_{2}}-{{E}_{1}})}{h}\]
B) slightly greater than \[\frac{({{E}_{2}}-{{E}_{1}})}{h}\]
C) slightly less than\[\frac{({{E}_{2}}-{{E}_{1}})}{h}\]
D) \[hv\]
Correct Answer: B
Solution :
If the kinetic energy of photoelectrons emitted from the metal surface is E^ and W is the work function of the metal, then from Einsteins equation of photoelectric effect, we have \[{{E}_{k}}=hv-W\] ...(i) Also, \[{{E}_{k}}={{E}_{2}}-{{E}_{1}}\] ...(ii) Hence, Eq. (i) is \[\frac{{{E}_{2}}-{{E}_{1}}}{h}+\frac{W}{h}=v\] Hence, frequency (v) is greater than\[\frac{{{E}_{2}}-{{E}_{1}}}{h}.\]
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