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Sloan [31]
3 years ago
15

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitte

d because sodium ions in the pickle become excited; their return to the ground state results in light emission. Calculate the energy gap between the excited and ground states for the sodium ion.
Physics
1 answer:
Alona [7]3 years ago
4 0

Answer:

2.11eV

Explanation:

We know that speed of light is it's wavelength times frequency.

\therefore f=v/\lambda\\=(3\times10^8m/s)/(589mm\times1m/1\times 10^9nm)\\=5.09\times10^1^4s^-1 \ or \ 5.09\times10^1^4Hz

Planck's constant is 6.626\times 10^3^4Js

The energy gap is calculated by multyplying the light's frequency by planck's constant:

E_c=5.09\times10^1^4s^-^1\times 6.626\times10^-^3^4Js\\\\=3.37\times 10^-^1^9J  \ \ \ \ \ \ #1eV=1.06\times 10^-^1^9J\\\\=2.11eV

Hence, the energy gap is 2.11eV

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True

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3 years ago
Read 2 more answers
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

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3 years ago
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Answer:

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