the de broglie wavelength of the electron ejected out when the radiation of energy 'E' strikes on the metal surface having threshold frequency equal to half of incident frequency is?
Answers
Given:
The radiation of energy 'E' strikes on the metal surface having threshold frequency equal to half of incident frequency.
To find:
De-Broglie wavelength of ejected electron.
Calculation:
Incident radiation has energy E , hence :
Let threshold value be E2 , hence :
From Equations (1) and (2) :
Now, applying Photo-Electric Effect Equation:
Now , we know that De-Broglie wavelength be :
So , final answer is :
Explanation:
Given:
The radiation of energy 'E' strikes on the metal surface having threshold frequency equal to half of incident frequency.
To find:
De-Broglie wavelength of ejected electron.
Calculation:
Incident radiation has energy E , hence :
\rm{E = hf \: \: \: .......(1)}E=hf.......(1)
Let threshold value be E2 , hence :
\rm{E2 = h \times ( \dfrac{f}{2}) \: \: \: .......(2)}E2=h×(
2
f
).......(2)
From Equations (1) and (2) :
\rm{\therefore \: E2 = \dfrac{E}{2} }∴E2=
2
E
Now, applying Photo-Electric Effect Equation:
\rm{KE = E - threshold \: energy}KE=E−thresholdenergy
\rm{ = > KE = E - E2}=>KE=E−E2
\rm{ = > KE = E - \dfrac{E}{2}}=>KE=E−
2
E
\rm{ = > KE = \dfrac{E}{2}}=>KE=
2
E
Now , we know that De-Broglie wavelength be :
\rm{ \therefore \: \lambda = \dfrac{h}{mv} }∴λ=
mv
h
\rm{ = > \: \lambda = \dfrac{h}{ \sqrt{2m(KE)} } }=>λ=
2m(KE)
h
\rm{ = > \: \lambda = \dfrac{h}{ \sqrt{2m( \frac{E}{2} )} } }=>λ=
2m(
2
E
)
h
\rm{ = > \: \lambda = \dfrac{h}{ \sqrt{mE} } }=>λ=
mE
h
So , final answer is :
\boxed{ \green{ \large{ \rm{ \: \lambda = \dfrac{h}{ \sqrt{mE} } }}}}
λ=
mE
h