A100 watt bulb emits monochromatic light of wavelength 400 nm. calculate the number of photons emitted per second by the bulb.

f = 7.5 E14 s-1

Explanation:

energy emitted by photon:

∴ λ = (400nm)(m/1 E9 nm) = 4.00 E-7 m

∴ h ( Planck's constant) = 6.626070150 E-34 J.s

∴ c (velocity of light) = 3.00 E8 m/s

∴ f (frequency): photon/s

If (1) = (2):

⇒ f = c / λ = E / h

⇒ f = (3.0 E8 m/s) / ( 4.00 E-7 m) = 7.5 E14 s-1

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.