Question

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant.
Wavelengths (nm)
500
450
400

$v \times 10^{-5}$($cms^{-1}$)
2.55
4.35
5.35

Answer 1

a) Let us suppose threshold wavelength to be ${ \lambda }_{ D }nm\quad (=\quad \lambda _{ 0 }\quad \times \quad 1{ 0 }^{ -9 }\quad m)$

The kinetic energy of the radiation is given as

$h(v\quad -\quad { v }_{ o })\quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

Or

$hc\quad \left( \frac { 1 }{ \lambda } \quad -\quad \frac { 1 }{ { \lambda }_{ D } } \right) \quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

$hc\left( \frac { 1 }{ 500\quad \times \quad 1{ 0 }^{ 9 } } \quad -\quad \frac { 1 }{ { \lambda }_{ 0 }\quad \times \quad 1{ 0 }^{ -9 } } \quad \right) \quad =\quad \frac { 1 }{ 2 } m\quad \left( 2.55\quad \times \quad 1{ 0 }^{ +5 }\quad \times \quad 1{ 0 }^{ -2 } \right)$

$\frac { hc }{ 1{ 0 }^{ -9 } } \left[ \frac { 1 }{ 500 } \quad -\quad \frac { 1 }{ { \lambda }_{ 0 } } \right] \quad =\quad \frac { 1 }{ 2 } m\quad { \left( 2.55\quad \times \quad 1{ 0 }^{ 3 } \right) }^{ 2 }\quad ..............(1)$

Similarly,

$\frac { hc }{ 1{ 0 }^{ -9 } } \left[ \frac { 1 }{ 450 } \quad -\quad \frac { 1 }{ { \lambda }_{ 0 } } \right] \quad =\quad \frac { 1 }{ 2 } m(3.45\quad \times \quad 1{ 0 }^{ 3 }{ ) }^{ 2 }..............(2)$

$\frac { hc }{ 1{ 0 }^{ -9 }m } \left[ \frac { 1 }{ 450 } \quad -\quad \frac { 1 }{ { \lambda }_{ 0 } } \right] \quad =\quad \frac { 1 }{ 2 } m(5.35\quad \times \quad 1{ 0 }^{ 3 }{ ) }^{ 2 }..............(3)$

Dividing equation (3) by equation (1)

$\frac { \left[ \frac { { \lambda }_{ o }-\quad 400 }{ 400{ \lambda }_{ o } } \right] }{ \left[ \frac { { \lambda }_{ o }\quad -\quad 500 }{ 500{ \lambda }_{ o } } \right] } \quad =\quad \frac { (5.35\quad \times \quad { 10 }^{ 3 }{ ) }^{ 2 } }{ (2.55\quad \times \quad { 10 }^{ 3 }{ ) }^{ 2 } }$

$\frac { 5{ \lambda }_{ o }\quad -\quad 2000 }{ 4{ \lambda }_{ o }\quad -\quad 2000 } \quad =\quad { \left( \frac { 5.35 }{ 2.55 } \right) }^{ 2 }\quad =\quad \frac { 28.6225 }{ 6.5025 }$

$\frac { 5{ \lambda }_{ o }\quad -\quad 2000 }{ 4{ \lambda }_{ o }\quad -\quad 2000 } \quad =\quad 4.40177$

$17.6070{ \lambda }_{ o }\quad -\quad 5{ \lambda }_{ o }\quad =\quad 8803.537\quad -\quad 2000$

${ \lambda }_{ o }\quad =\quad \frac { 68055.537 }{ 12.607 }$

${ \lambda }_{ o }\quad =\quad 539.8\quad nm$

${ \lambda }_{ o }\quad =\quad 540\quad nm$

Threshold wavelength $({ \lambda }_{ D }) = 540 nm$

Substituting this value in equation (3), we get

$=\quad h\quad \times \quad \frac { (3\quad \times { \quad 10 }^{ 8 }) }{ { 10 }^{ -9 } } \left[ \frac { 1 }{ 400 } \quad -\quad \frac { 1 }{ 540 } \right]$

$=\quad \frac { [(9.11\quad \times \quad { 10 }^{ -31 })(5.20\quad \times \quad { 10 }^{ 6 })] }{ 2 } \quad =\quad 6.66\quad \times \quad 1{ 0 }^{ -34 }\quad Js$