Question

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The
maximum speeds of the photoelectrons corresponding to these wavelengths are ph and fly respectively,
If the ratio nearly : th= 4:1 and hc = 1240 eV nm, the work function of the metal is​

Answer 1

The work function of the metal is​ 3.9 eV.

Explanation:

=> Here, a metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm.

Ratio u₁/u₂ = 4/1 [∴ u₁²/u₂² = 16/1]

hc = 1240 eV nm

=> According to the formula of kinetic energy:

KEmax = hc/λ - w

=> By substituting the value in above formula, we get

For light of 248 nm wavelengths:  

1/2 mu₁² = hc/μ₁ - w

1/2 mu₁² = 1240/248 - w

1/2 mu₁² = 5ev - w ...(1)

For light of 310 nm wavelengths:  

1/2 mu₂² = hc/μ₂ - w

1/2 mu₂² = 1240/310 - w

1/2 mu₂² = 4ev - w ...(2)

=> Dividing eq(1) by eq(2), we get

$\frac{1/2mu1^2}{1/2mu2^2}=\frac{5ev - w}{4ev - w}$

$\frac{u1^2}{u2^2} =\frac{5 - w}{4 - w}$

16 = $\frac{5-w}{4 - w}$

16(4-w) = 5 - w

64 - 16w = 5 - w

64 - 5 = 16w - w

59 = 15w

w = 59/15

w = 3.93 eV ≈ 3.9 eV

Thus, the work function of the metal is 3.9 eV.

Learn more:

Q:1 The work function of a metal is 4.2ev . If radiations of 2000 amsrtong fall on the metal, then find the kinetic energy of fastest photon electron.

Click here: https://brainly.in/question/10825615

Q:2 Work function of a metal is 2.3eV. Calculate the threshold frequency of the metal.

Click here: https://brainly.in/question/6331203

Answer 2

Given :

• A metal surface is illuminated by light of two wavelengths $\sf{\lambda_1=248nm\:and\:\lambda_2=310nm}$
• Maximum speeds of the photoelectrons corresponding to these wavelength are$\sf{u_1\:\:and\:\:u_2}$
• Ratio of $\sf{u_1:u_2=4:1}$

To Find :

The work function of the metal , given hc = 1240 eVnm

Theory:

• Einstein's Photoelectric Equation :

$\rm\blue{h\:\nu=W+KE_{max}}$

and $\rm\green{KE_{max}=eV_s}$

Then,

$\bf\purple{h\:\nu=W+eV_s}$

Where,

h = plank constant

W = Work function

$\sf{eV_{s}=}$ Stopping potential

f = frequency

• Work Function :

The minimum energy required for emission of electrons from metal is called work function.

$\sf{W=h\:\nu_o}$

Where, $\sf{\nu_o}$ is the threshold frequency

Solution :

Let the work function of the metal be W .

Given : Ratio $\sf{u_1:u_2=4:1}$

$\sf\implies{u_1=4u_2}$

Case -1 :

When $\sf{Wavelength,\lambda_1=248nm}$

and $\sf{Velocity,u_1}$

Then, by Einstin's Photoelectric Equation:

$\sf{\dfrac{hc}{\lambda_1}=W+\dfrac{1}{2}mu^2_{1}}$

Put the given values

$\sf\implies{\dfrac{1240}{248}=W+\dfrac{1}{2}m(4u_2)^2}$

$\sf\implies{5eV=W+\dfrac{1}{2}m(16u^2_2)}$

$\sf\implies{\dfrac{1}{2}m(16u^2_2)=5-W}$ equation (1)

Case -2:

When $\sf{Wavelength,\lambda_2=310nm}$

and $\sf{Velocity,u_2}$

Then, by Einstin's Photoelectric Equation:

$\sf{\dfrac{hc}{\lambda_2}=W+\dfrac{1}{2}mu^2_2}$

Put the given values

$\sf\implies{\dfrac{1240}{310}=W+\dfrac{1}{2}m(u_2)^2}$

$\sf\implies{4eV=W+\dfrac{1}{2}m(u_2)^2}$

$\sf\implies{\dfrac{1}{2}m(u^2_2)=4-W}$ equation (2)

Now Divide Equation (1) and (2) , then

$\sf\implies{\dfrac{\frac{1}{2}m(16u^2_2)}{\frac{1}{2}mu^2_2}=\dfrac{5-W}{4-W}}$

$\sf\implies{16=\dfrac{5-W}{4-W}}$

$\sf\implies{16(4-W)=5-W}$

$\sf\implies{64-16W=5-W}$

$\sf\implies{16W-W=64-5}$

$\sf\implies{15W=59}$

$\sf\implies{W=\dfrac{59}{15}}$

$\sf\implies{W=3.933eV}$

$\sf\implies{W\:\approx3.9eV}$

Therefore, The work function of the metal is 3.9 eV (approx)