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Question: Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and cause the emission of photoel...

Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and cause the emission of photoelectrons for which the stopping potential is 0.5V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5V . with this data, The value of

A

4×10−59Vs4 \times 10 ^ { - 59 } \mathrm { Vs }

B

0.25×10−15Vs0.25 \times 10 ^ { - 15 } \mathrm { Vs }

C

4×10−15Vs4 \times 10 ^ { - 15 } \mathrm { Vs }

D

Answer

4×10−15Vs4 \times 10 ^ { - 15 } \mathrm { Vs }

Explanation

Solution

: Here eV=hcλ−W0e V = \frac { h c } { \lambda } - W _ { 0 }

⇒0.5=he(c6×10−7)−W0e\Rightarrow 0.5 = \frac { h } { e } \left( \frac { c } { 6 \times 10 ^ { - 7 } } \right) - \frac { W _ { 0 } } { e }…… (i)

Similarly, 1.5=he(c4×10−7)−W0e1.5 = \frac { h } { e } \left( \frac { c } { 4 \times 10 ^ { - 7 } } \right) - \frac { W _ { 0 } } { e }….. (ii)

From equation (i) and (ii)

1=hec10−7[14−16]⇒he=12×10−73×108=4×10−15vs1 = \frac { h } { e } \frac { c } { 10 ^ { - 7 } } \left[ \frac { 1 } { 4 } - \frac { 1 } { 6 } \right] \Rightarrow \frac { h } { e } = \frac { 12 \times 10 ^ { - 7 } } { 3 \times 10 ^ { 8 } } = 4 \times 10 ^ { - 15 } v s