Answer: C. Metals are found on the left side of the periodic table.
Explanation:
The periodic table contains vertical columns called as groups and horizontal rows called as periods.
Period 2 contains 8 elements which are lithium, beryllium , boron , carbon, nitrogen, oxygen , fluorine and neon. Only Lithium and beryllium are metals.
Group 18 contains all the noble gases which are all non metals.
Metals are the elements which loose electrons easily and form positive ions. Non-metals are the elements which can gain electrons easily and form negative ions.
Metals are present on left side of the periodic table and as we move to right side of the periodic table , the metallic character decreases and thus non metals are found on the right side of the periodic table.
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
Next, you use the equation (2) and solve for λ:
C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
Answer:
0.438kg/ms-¹
Explanation:
Momentum, denoted by p, can be calculated by using the formula;
p = mv
Where;
m = mass (kg)
v = velocity (m/s)
Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹
Momentum (p) of crawling baby = 7.29 kg kg × 0.234 m/s = 1.706kg/ms-¹
Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:
{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V
