Answer:
Calcium (Ca) and Scandinavium (Sc)
Explanation:
They are all on the same period.
Answer:
There are many different types of preservatives like Benzoic acid, Calcium Sorbate, Erythorbic Acid, Potassium Nitrate and Sodium Benzoate. Some act like antioxidants used for slowing down spoilage like Ascorbyl Palmitate, Butylated Hydroxy anisole (BHA) and Butylated Hydroxytoluene (BHT
Answer:
energy
Explanation:
The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.
The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.
The process is called excitation and de-excitation.
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits. For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum
For this problem we use the wave equation. It is expressed as the speed (c) is equal to the product of frequency (f) and wavelength (v).
c = v x f
We know the wavelength of the an red light which is 6.5 x 10^-7 m. Now, we solve for the wavelength of the unknown wave to see the relation between the two waves.
2.998 X 10^8 = 5.3 X 10^15 X v
v = 2.998 X 10^8 / (5.3 X 10^15) = 5.657 X 10^-8 m
Therefore, the wavelength of the unknown wave is less than the wavelength of the red light.
