<u>Explanation:</u>
<u>1. Have many moons:</u>
- Jupiter is the fifth planet from the Sun and the biggest in the Solar System and it has fifty-three moons which are confirmed and twenty-six provisional moons and totally it has seventy-nine moons and it is the only planet which has many moons.
<u>2. Have a rocky composition:</u>
- The planets which have rocky composition are also called the terrestrial planets.
- The planets which have rocky composition are listed below mercury, venus, earth, and mars and they are smaller in size.
<u>3. Revolve quickly around the Sun: </u>
- Mercury is the quickest planet, which rushes around the sun at 47.87 km/s. And it revolves around the sun quickly.
<u> 4. Rotate quickly on their axes: </u>
- The giant gas planets like Jupiter, Saturn, etc... spin more quickly on their axes than the other planets
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.
The mass number is the summation of number of proton and neutron present in a nucleus of an atom. For the neutral atom the number of positive charge (number of proton) must be equal to the number of electrons. The number of electrons present in an atom is the atomic number of the atom. The standard way to express the mass number (a) and atomic number (m) of a atom (say X) is 
. Now for silicon number of electron or atomic number is 14. And the mass number (a) given 29. Thus the expression nucleus of silicon will be 
