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# A photon of wavelength 4.0 x $10^{-7}$m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate:i. The energy of the photon (eV).ii. The kinetic energy of the emission.Iii. The velocity of the photoelectron.

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I) E = hv/wavelength =(6.626*10^-34)(3*10^8)/4*10^-7 = 4.9695*10^-19J ii)K.E=4.9695*10^-19/1.60*10^-19 =3.105eV

1. 4.97 ×10-19 / 1.602 ×10-19 eV 2. 0.97eV 3. 5084×105 ms-1

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Energy of photon= hc/λ h=plank constant = 6.63* h*c= 1240 eV λ= wavelength of light E= / J E=3.102 eV 2.)Kinetic energy of emission= energy of photon- work function =3.102-2.13 eV = 0.972 eV 3.) kinetic energy of... read more

Energy of photon= hc/λ

h=plank constant = 6.63*$10^{-34}$

h*c= 1240 eV

λ= wavelength of light

E= $6.262\times10^{-34}$$\times&space;3\times&space;10^{8}$/$4\times&space;10^{-7}$ J

E=3.102 eV

2.)Kinetic energy of emission= energy of photon- work function

=3.102-2.13 eV

= 0.972 eV

3.) kinetic energy of electron = 1/2*m*$v^{2}$ = 0.972 eV

v=$\sqrt{2\times&space;0.963\times&space;1.6\times&space;10^{-19}/9.1*10^{-31}$

=5.84$\times&space;10^{^{5}}$ m/s

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I) E = hv/wavelength =(6.626*10^-34)(3*10^8)/4*10^-7 = 4.969*10^-19 J

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Energy of photon= hc/λ h=plank constant = 6.63* c=speed of light=3* h*c= 1240 eV λ= wavelength of light E= / joule E=3.102 eV 2.)Kinetic energy of emission= energy of photon- work function =3.102-2.13 eV ... read more

Energy of photon= hc/λ

h=plank constant = 6.63*$10^{-34}$

c=speed of light=3*$10^{^{8}}$

h*c= 1240 eV

λ= wavelength of light

E= $6.262\times10^{-34}$$\times&space;3\times&space;10^{8}$/$4\times&space;10^{-7}$ joule

E=3.102 eV

2.)Kinetic energy of emission= energy of photon- work function

=3.102-2.13 eV

= 0.972 eV

3.)Velocity of free Electrons

kinetic energy of electron = 1/2*m*$v^{2}$ = 0.972 eV

v=$\sqrt{2\times&space;0.963\times&space;1.6\times&space;10^{-19}/9.1*10^{-31}$

=5.84$\times&space;10^{^{5}}$ m/s

Tutor

E=hc/wavelength=3.10ev K.E. of photoelectron = (KE of incident photon) - (Work Function of Metal)=3.10-2.13=0.87ev Velocity of photoelctron={2*k.E/m}^{0.5}=30um/sec

Tutor

(1) Energy of the photon (E) = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/4×10-7 m = 4.97×10-19 J= 4.97×10-19/1.602×10-19 eV (2) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV (3) 1/2 mv2 = 0.97 eV = 0.97×1.602×10-19... read more

(1) Energy of the photon (E) = hν = hc/λ = (6.626×10-34 Js×3.0×10ms-1)/4×10-7 m = 4.97×10-19 J
= 4.97×10-19/1.602×10-19 eV

(2) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV

(3) 1/2 mv= 0.97 eV = 0.97×1.602×10-19 J
⇒ 1/2×(9.11×10-31 kg)×v= 0.97×1.602×10-19 J
⇒ v= 0.341×1012 = 34.1×1010
⇒ v = 5.84×10ms-1

First of all we have to find the enenry(E) of Photon by using the Formula E=hc/λ here 'h' is the Plank's constant. 'c' is the Speed of light and λ is the wavelength Now we have to think what is the term "WORK FUNCTION" used in the Question WORK FUNCTION is the MINIMUM amount of... read more

First of all we have to find the enenry(E) of Photon by using the Formula

E=hc/λ

here 'h' is the Plank's constant.

'c' is the Speed of light

and λ is the wavelength

Now we have to think what is the term "WORK FUNCTION" used in the Question

WORK FUNCTION is the MINIMUM amount of energy required to release a ELECTRON from the Metal

It is given in the Question

Now we are able to give the answer of the 2nd part of the Question

i.e Kinetic energy of emission== Total energy of photon- Energy required for emission(work function)

Now the question is WHAT IS THE VELOCITY OF PHOTOELECTRON??

From 2nd part of Question We know the Kinetic Energy of Photoelectron.

Use the formula,

K.E= ½*mass*(velocity)²

for the value of "mass of electron" , "plank's constant" and "velocity of light "use your book.

THANK YOU

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