JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 19)
What is the work function of the metal if the light of wavelength 4000$$\mathop A\limits^ \circ $$ generates photoelectrons of velocity 6 $$ \times $$ 105 ms–1 from it ?
(Mass of electron = 9 $$ \times $$ 10–31 kg;
Velocity of light = 3 $$ \times $$ 108 ms$$-$$1
Plank's constant = 6.626 $$ \times $$ 10–34 Js;
Charge of electron = 1.6 $$ \times $$10–19 JeV–1)
(Mass of electron = 9 $$ \times $$ 10–31 kg;
Velocity of light = 3 $$ \times $$ 108 ms$$-$$1
Plank's constant = 6.626 $$ \times $$ 10–34 Js;
Charge of electron = 1.6 $$ \times $$10–19 JeV–1)
4.0 eV
0.9 eV
2.1 eV
3.1 eV
Explanation
E = $$\phi $$ + K.E
$$ \Rightarrow $$ h$$\nu $$ = $$\phi $$ + $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $$\phi $$ = $$h\nu - {1 \over 2}m{v^2}$$
= $${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$$
= 3.35 $$ \times $$ 10-19 J
$$ \Rightarrow $$ $$\phi $$ = $${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV
= 2.0934 eV $$ \simeq $$ 2.1 eV
$$ \Rightarrow $$ h$$\nu $$ = $$\phi $$ + $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $$\phi $$ = $$h\nu - {1 \over 2}m{v^2}$$
= $${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$$
= 3.35 $$ \times $$ 10-19 J
$$ \Rightarrow $$ $$\phi $$ = $${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV
= 2.0934 eV $$ \simeq $$ 2.1 eV
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