In a photo-electric cell, current stops when a negative potential of 0.5 V is given to the collector with respect to the emitter. The maximum kinetic energy of the emitted electrons is :

0.8 \times 10^{- 19} J

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0.8 \times 10^{- 19} e r g

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0.5 J

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0.5 e r g

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Solution

The correct option is B $0.8 \times 10^{- 19} J$
$s t o p p i n g p o t e n t i a l = (\frac{h c}{λ} - ϕ)$ ........(1)

$K . E_{m a x} = \frac{h c}{λ} - ϕ$ -.......2)

so, From (1) and (2)

$\frac{K . E_{m a x}}{e} = s t o p p i n g p o t e n t i a l$

$K . E_{m a x} = e \times s t o p p i n g p o t e n t i a l$

= $1.6 \times 10^{- 19} \times 0.5$

= $0.8 \times 10^{- 19} J$

So, the answer is option (A).

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In a photo-electric cell, current stops when a negative potential of 0.5 V is given to the collector with respect to the emitter. The maximum kinetic energy of the emitted electrons is :

The correct option is B 0.8×10−19Jstopping potential=(hcλ−ϕ)........(1)K.Emax=hcλ−ϕ -.......2)so, From (1) and (2)K.Emaxe=stopping potentialK.Emax=e×stopping potential= 1.6×10−19×0.5= 0.8×10−19JSo, the answer is option (A).