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Volgvan
3 years ago
5

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

energy of the electron plus the kinetic energy of the emitted electron. When photons of 4.00×10-7 m light strike a calcium metal surface, electrons are ejected with a kinetic energy of 6.26×10-20 J. a Calculate the binding energy of the calcium electrons.
Chemistry
1 answer:
slega [8]3 years ago
5 0

Answer:

Binding\ energy=43.43\times 10^{-20}\ J

Explanation:

Using the expression for the photoelectric effect as:

E=h\nu_0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

\nu_0=\frac {c}{\lambda_0}

Applying the equation as:

\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

Given, \lambda=4.00\times 10^{-7}\ m

\frac {hc}{\lambda_0} is the binding energy or threshold energy

\frac {1}{2}\times m\times v^2 is the kinetic energy of the electron emitted.  = 6.26\times 10^{-20}\ J

Thus, applying values as:

\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy

\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J

Binding\ energy=43.43\times 10^{-20}\ J

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Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


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Explanation:

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At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha
Alekssandra [29.7K]

Explanation:

Density =\frac{Mass}[Volume}

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball =V=560 cm^3=560 ml=0.560 L

D =\frac{0.12 g}{0.560 L}=0.214 g/L

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = d_1=1.830 g/L

Mass of the carbon dioxide gas :

1.830 g/L\times 0.560 L=1.0248 g

Total density of filled ball with carbon dioxide gas:

\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = d_2=0.0899 g/L

Mass of the hydrogen gas :

1.830 g/L\times 0.560 L=0.050344 g

Total density of filled ball with hydrogen gas:

\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = d_3=1.330 g/L

Mass of the oxygen gas :

1.330 g/L\times 0.560 L=1.7448 g

Total density of filled ball with oxygen gas:

\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = d_4=1.165 g/L

Mass of the nitrogen gas :

1.165 g/L\times 0.560 L=0.6524 g

Total density of filled ball with nitrogen gas:

\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

\frac{0.12g +m}{0.560L}>1.189 g/L

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0
2 years ago
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