Question

When light of wavelength 470 nm falls on the surface of potassium metal electrons are emitted with the velocity of 6.4 ×10^4 metre per second what is the minimum energy required per mole to remove an electron

"Please provide solution and also tell me from which book the question is taken "​

Answer 1

Answer:

$\displaystyle{W_0=4.219\times10^{-15} \ J}$

Explanation:

From photoelectric effect we have studied that when a light ( h ν )  falls on metal surface it absorb some energy ( Wₙ ) and reflect back in form of kinetic energy ( 1 / 2 m v² ) .

From here we get equation :

h ν  = W + 1 / 2 m v²

ν = c / λ  

h  c / λ = W + 1 / 2 m v²

W  = h  c / λ - 1 / 2 m v²

Given :

λ  = 470 nm

v = 6.4 × 10⁴ m / sec .    

We know value of  [  c , h and m ]

Now putting all values now :

$\displaystyle{W_0=\left[6.626\times10^{-34}\times\dfrac{3\times10^8}{470\times10^{-9}} -\dfrac{1}{2}\times9.1\times10^{-31}\times\left(6.4\times10^{4}\right)^2\right]}\\\\\\\displaystyle{W_0=4.220\times10^{-15}-1.86\times10^{-21}}\\\\\\\displaystyle{W_0=10^{-15}\left(4.22-1.86\times10^{-6\right)}}\\\\\\\displaystyle{W_0=10^{-15}\left(4.219\right)}\\\\\\\displaystyle{W_0=4.219\times10^{-15} \ J}$

Hence we get answer .

Answer 2

Answer:

253.6kJ mol-1

Explanation:

Kinetic energy of photon electrons =12mv2

=12×9.1×10−31×(6.4×104)2=1.864×10−21J we know that,

Absorbed energy from light = Threshold energy + kinetic energy of photoelectrons

hcλ=W0+12mv2

⇒6.626×10−34×3×108470×10−9=W0+1.864×10−21J

⇒W0=419.206×10−21J

Energy required to eject one moles of electrons

=419.21×10−21×6.023×1023

=252.4×103Jmol−1=252.4kJmol−1