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Marysya12 [62]
3 years ago
13

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

he work function W0 of this metal? Express your answer in electron volts.
Physics
1 answer:
spin [16.1K]3 years ago
4 0

Answer:

2.7067 eV

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

\lambda_0 = Threshold wavelength = 459 nm

Work function is given by

W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV

The work function W0 of this metal is 2.7067 eV

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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Stels [109]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 years ago
An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit
zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

F=qE

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

q=3.0\cdot 10^{-6}C

E=-1.5\cdot 10^4 N/C (negative because inward)

So the strength of the electrostatic force is

F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0
3 years ago
Read 2 more answers
In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required
konstantin123 [22]

Answer:

2.46 eV

Explanation:

It is given that,

The energy of light that fall on the metal = 3.56 eV

The stopping potential of the metal = 1.1 V

We need to find the work function of the metal. It is given by the relation as follows :

W = E-KE ...(1)

Where KE is the kinetic energy of the ejected electron and it is given by :

KE = V×e

= 1.1 eV

Put all the values in formula (1)

W = 3.56 eV - 1.1 eV

= 2.46 eV

Hence, the work function of the metal is 2.46 eV. Hence, the correct option is (c).

5 0
2 years ago
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