Question

The critical wavelength for producing the photoelectric effect in tungsten is $ 2600{\text{ }}{{\text{A}}^ \circ } $ . What will be the wavelength necessary to produce photoelectrons from tungsten having twice the K.E of those produced at $ 2200{\text{ }}{{\text{A}}^ \circ } $ .
 $ (i){\text{ }}2900{\text{ }}{{\text{A}}^ \circ } $
 $ (ii){\text{ 30}}00{\text{ }}{{\text{A}}^ \circ } $
 $ (iii){\text{ 19}}00{\text{ }}{{\text{A}}^ \circ } $
 $ (iv){\text{ 35}}00{\text{ }}{{\text{A}}^ \circ } $

Answer 1

Hint: The critical wavelength of tungsten is also known as the threshold wavelength of material. Thus with the help of threshold wavelength we will find the Kinetic energy of photons. Then we will double this kinetic energy and then we will find the wavelength of photons which will produce that much amount of energy.
 $ hv{\text{ = }}h{v_ \circ }{\text{ + K}}{\text{.E}} $
Here, $ hv $ is the energy supplied to material, $ h{v_ \circ } $ is the threshold energy of electrons and K.E represents the kinetic energy of electrons.

Complete answer:
According to photoelectric effect the energy $ \left( {hv} \right) $ which is supplied to material is used for overcome the threshold energy of electrons $ \left( {h{v_ \circ }} \right) $ and providing kinetic energy to the electrons. Thus this energy can be represented as,
 $ hv{\text{ = }}h{v_ \circ }{\text{ + K}}{\text{.E}} $
We know that, $ hv{\text{ = }}\dfrac{{hc}}{\lambda } $ , therefore the above equation can be deduced as,
 $ \dfrac{{hc}}{\lambda }{\text{ = }}\dfrac{{hc}}{{{\lambda _ \circ }}}{\text{ + K}}{\text{.E}} $
Here $ \lambda $ is known as the wavelength of photoelectrons which is supplied to tungsten and $ {\lambda _ \circ } $ is the critical or threshold wavelength of tungsten material. Since we are given that critical wavelength of tungsten is $ 2600{\text{ }}{{\text{A}}^ \circ } $ and wavelength of photoelectrons is $ 2200{\text{ }}{{\text{A}}^ \circ } $ therefore the equation can be deduced as,
 $ \dfrac{{hc}}{{2200}}{\text{ = }}\dfrac{{hc}}{{2600}}{\text{ + K}}{\text{.E}} $
 $ {\text{K}}{\text{.E = }}\dfrac{{hc}}{{2200}}{\text{ - }}\dfrac{{hc}}{{2600}} $ ________________ $ {\text{(1)}} $
Now we have to find the wavelength of photoelectrons which can have twice the energy as produced at $ 2200{\text{ }}{{\text{A}}^ \circ } $ . Therefore now new kinetic energy will be $ {\text{2K}}{\text{.E}} $ . Hence we can find the wavelength of such photoelectrons as,
 $ \dfrac{{hc}}{\lambda }{\text{ = }}\dfrac{{hc}}{{{\lambda _ \circ }}}{\text{ + K}}{\text{.E}} $
Now kinetic energy will be $ {\text{2K}}{\text{.E}} $ , therefore
 $ \dfrac{{hc}}{\lambda }{\text{ = }}\dfrac{{hc}}{{{\lambda _ \circ }}}{\text{ + 2K}}{\text{.E}} $
Now again using $ {\lambda _ \circ }{\text{ = }}2600{\text{ }}{{\text{A}}^ \circ } $ and using value of kinetic energy from equation $ {\text{(1)}} $ , we get the result as,
 $ \dfrac{{hc}}{\lambda }{\text{ = }}\dfrac{{hc}}{{2600}}{\text{ + 2}}\left( {\dfrac{{hc}}{{2200}}{\text{ - }}\dfrac{{hc}}{{2600}}} \right) $
Taking $ hc $ common from both side and the equation will become as,
 $ \dfrac{1}{\lambda }{\text{ = }}\dfrac{1}{{2600}}{\text{ + 2}}\left( {\dfrac{1}{{2200}}{\text{ - }}\dfrac{1}{{2600}}} \right) $
Now solving the equation we will obtained wavelength of photoelectrons as,
 $ \dfrac{1}{\lambda }{\text{ = }}\dfrac{2}{{2200}}{\text{ - }}\dfrac{1}{{2600}} $
 $ \lambda {\text{ = 1906}}{\text{.66 }}{{\text{A}}^ \circ } $
Thus the wavelength of photoelectrons required is $ {\text{19}}00{\text{ }}{{\text{A}}^ \circ } $ .
Hence the correct option is $ (iii){\text{ 19}}00{\text{ }}{{\text{A}}^ \circ } $ .

Note:
We can also find the value of kinetic energy in term of wavelength only and then we will find the ratio of $ {\text{K}}{\text{.E}} $ and $ {\text{2K}}{\text{.E}} $ to find the wavelength of photoelectrons. The answer will be the same in both cases. We do not need to calculate the kinetic energy in each by putting values of $ h $ and $ c $ as they will cancel out at the end.