Question

The energy required to remove an electron from the surface of Na metal is 2.3 eV. What is the largest wavelength of radiation which can show the photoelectric effect?

Answer 1

"Photoelectric effect:

The phenomenon of emission from a metallic surface by use of light energy is called photoelectric effect.  

From the given,

Energy required to show photo emission$=\quad 2.3\quad ev\quad =\quad 2.3\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 }\quad J$

Energy formula, $E\quad =\quad \frac { hc }{ \lambda  }$

$h\quad =\quad 6.626\quad \times { \quad 10 }^{ -34 }$

$c\quad =\quad 3\quad \times \quad { 10 }^{ 8 }\quad m/s$

Substitute the values

$\lambda \quad =\quad \frac { hc }{ E }$

$=\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad \times \quad 3\quad \times { \quad 10 }^{ 8 } }{ 2.3\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 } } \quad =\quad 5.4\quad \times \quad { 10 }^{ -7 }\quad m$"

Answer 2

"Photoelectric effect:

The phenomenon of emission from a metallic surface by use of light energy is called photoelectric effect.

From the given,

Energy required to show photo emission=\quad 2.3\quad ev\quad =\quad 2.3\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 }\quad J=2.3ev=2.3×1.6×10

−19

J

Energy formula,

h\quad =\quad 6.626\quad \times { \quad 10 }^{ -34 }h=6.626×10

−34

c\quad =\quad 3\quad \times \quad { 10 }^{ 8 }\quad m/sc=3×10

8

m/s

Substitute the values

\lambda \quad =\quad \frac { hc }{ E }λ=

E

hc

=\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad \times \quad 3\quad \times { \quad 10 }^{ 8 } }{ 2.3\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 } } \quad =\quad 5.4\quad \times \quad { 10 }^{ -7 }\quad m=

2.3×1.6×10

−19

6.626×10

−34

×3×10

8

=5.4×10

−7

m "