Solution: Maximum kinetic energy, K Emax=hcλincident-hcλthreshold=hc1λincident-1λthreshold =12401155-1310 eV=12401155-1310 eV =4 eV,So, e×Stopping potential=4 eV⇒Stopping potential=4 V

Phy 21. Que

Solution:
Maximum kinetic energy,

K . E_{m a x} = \frac{h c}{λ_{i n c i d e n t}} - \frac{h c}{λ_{t h r e s h o l d}} = h c (\frac{1}{λ_{i n c i d e n t}} - \frac{1}{λ_{t h r e s h o l d}}) = 1240 (\frac{1}{155} - \frac{1}{310}) eV = 1240 (\frac{1}{155} - \frac{1}{310}) eV = 4 eV, So, e \times S topping potential = 4 eV \Rightarrow Stopping potential = 4 V

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