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STatiana [176]
3 years ago
10

Complete and balance the following nuclear equations by including the missing particle. You may want to reference (Pages 906 - 9

09) Section 21.2 while completing this problem. Part A 25298Cf+105B → 310n+?
Chemistry
1 answer:
Alja [10]3 years ago
4 0

Answer:

The missing particle is _{103}^{259}\textrm{Lr}.

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

^{252}_{98}\textrm{Cf}+^{10}_{5}\textrm{B}\rightarrow ^A_Z\textrm{X}+3^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

252 +10 = A+ 3

A = 259

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

98 +5 = Z + 0

Z = 103

Hence, the isotopic symbol of unknown element is _{103}^{259}\textrm{Lr}

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3 years ago
3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv
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Answer:

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

8 0
2 years ago
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