What is work function of metal?

Chemistry Level 2

When light of frequency 8.6 × 1 0 14 8.6 \times 10^{14} Hz is incident on metal surface area, the maximum kinetic energy of photoelectrons is 0.500 eV. What is the work function of metal?

3.5 eV 3.8 eV 3.1 eV 4.3 eV 4.1 eV 0.5 eV

Winod Dhamnekar by Winod DHAMNEKAR , 55, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Einstein's photoelectric equation -

h ν = K . E . p h o t o e l e c t r o n + W 0 h\nu=K.E._{photoelectron}+W_0 , where W 0 W_0 is the work function of the target, h = 6.627 × 1 0 − 34 h=6.627\times 10^{-34} J-s is the Planck's constant. Given ν = 8.6 × 1 0 14 \nu=8.6\times 10^{14} Hz. So h ν = 56.9922 × 1 0 − 20 h\nu=56.9922\times 10^{-20} Joules = 56.9922 × 1 0 − 20 1.59 × 1 0 − 19 ≈ 3.562 =\dfrac{56.9922\times 10^{-20}}{1.59\times 10^{-19}}\approx 3.562 eV. Therefore W 0 = 3.562 − 0.500 = 3.062 ≈ 3.1 W_0=3.562-0.500=3.062\approx \boxed {3.1} eV.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Winod Dhamnekar
Mar 16, 2020

K E m a x = h f − ϕ ⇒ ϕ = ( 4.14 × 1 0 − 15 e V s ∗ 8.6 × 1 0 14 KE_{max}=hf-\phi \Rightarrow \phi=(4.14\times 10^{-15} eVs* 8.6\times 10^{14} Hz s − 1 ) − 0.5 e V s = 3.1 e V s s^{-1})-0.5 eVs=3.1 eVs

0 Helpful 0 Interesting 0 Brilliant 0 Confused

this question is of mains 2020 na?

VANSH SARDANA - 1 year, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...