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4 years ago

An electron with a kinetic energy of 3.90 ev collides with a sodium atom.

1 answer:

8 0

The wavelength transition line is
Δ = hc/Δ∈
=1240 e V. nm/3.75e V - 0,0ev
=330 .nm
Another wavelength transisiton is
Δ = hc/Δ∈
1240e V. nm/3.19e V -2.104e V
1140 .nm
Some other wavelength transition are;
590, 816, 2180, 8860.

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3 years ago

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3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

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3 years ago