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kicyunya [14]
2 years ago
11

A helium laser emits light with a wavelength of 633 nm. What is the frequency, v, of the light?​

Chemistry
1 answer:
Elanso [62]2 years ago
8 0

Answer:

4.74×10¹⁴ Hz.

Explanation:

From the question given above, the following data were obtained:

Wavelength of light = 633 nm.

Frequency of light =?

Next, we shall convert 633 nm to metre (m)

This can be obtained as follow:

1×10⁹ nm = 1 m

Therefore,

633 nm = 633 nm × 1 m / 1×10⁹ nm

633 nm = 6.33×10¯⁷ m

Thus, 633 nm is equivalent to 6.33×10¯⁷ m

Finally, we shall determine the frequency of the light as follow:

Wavelength of light = 6.33×10¯⁷ m

Velocity of light = 3×10⁸ m/s

Frequency of light =?

Velocity = wavelength x frequency

3×10⁸ = 6.33×10¯⁷ x frequency

Divide both side by 6.33×10¯⁷

Frequency = 3×10⁸ / 6.33×10¯⁷

Frequency = 4.74×10¹⁴ Hz.

Therefore, the frequency of the light is 4.74×10¹⁴ Hz.

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\bold{\huge{\orange{\underline{ Solution}}}}

\bold{\underline{ Given :- }}

  • <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
  • <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

\bold{\underline{ To \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

\bold{\underline{ Let's \:Begin:- }}

<u>We </u><u>know </u><u>that</u><u>, </u>

Amount of heat energy = mass * specific heat * change in temperature

<u>That </u><u>is, </u>

\sf{\red{ Q = mcΔT }}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

\sf{ Q = 250 × 4.180 ×(0 - 100 )}

\sf{ Q = 250 × 4.180 × - 100 }

\sf{ Q = 250 × - 418}

\sf{\pink{ Q = - 104,500 J }}

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

\bold{\underline{ Now :- }}

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic

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  • <u>Endothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>absorbed </u><u>.</u>
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