JEE MAIN - Chemistry (2020 - 2nd September Evening Slot - No. 3)
The work function of sodium metal is
4.41 $$ \times $$ 10–19 J. If photons of wavelength 300 nm
are incident on the metal, the kinetic energy of
the ejected electrons will be (h = 6.63 $$ \times $$ 10–34 J s;
c = 3 $$ \times $$ 108 m/s) ________ × 10–21 J.
Answer
222
Explanation
E = W + KEmax
$$ \Rightarrow $$ KEmax = E - W
= $${{hc} \over \lambda }$$ - 4.41 $$ \times $$ 10–19
= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {300 \times {{10}^{ - 9}}}}$$ - 4.41 $$ \times $$ 10–19
= 2.22 × 10–19 J
= 222 × 10–21 J
$$ \Rightarrow $$ KEmax = E - W
= $${{hc} \over \lambda }$$ - 4.41 $$ \times $$ 10–19
= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {300 \times {{10}^{ - 9}}}}$$ - 4.41 $$ \times $$ 10–19
= 2.22 × 10–19 J
= 222 × 10–21 J
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