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Usimov [2.4K]
2 years ago
10

X-rays had a frequency of about 3x10^18 determine it's wavelength (in nm) and energy per photon

Chemistry
1 answer:
Serggg [28]2 years ago
7 0

Answer:

E = 19.89×10⁻¹⁶ J

λ = 1×10⁻¹ nm

Explanation:

Given data:

Frequency of xray = 3×10¹⁸ Hz

Wavelength of xray = ?

Energy of xray = ?

Solution:

speed of wave = wavelength × frequency

speed = 3×10⁸ m/s

3×10⁸ m/s  = λ ×3×10¹⁸ s⁻¹

λ = 3×10⁸ m/s  / 3×10¹⁸ s⁻¹

λ = 1×10⁻¹⁰m

m to nm:

λ = 1×10⁻¹⁰m×10⁹

λ = 1×10⁻¹ nm

Energy of x-ray:

E = h.f

h = plancks constant = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js ×3×10¹⁸ s⁻ ¹

E = 19.89×10⁻¹⁶ J

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A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
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Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

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