In his experiment, Balmer introduced hydrogen gas in a discharged tube in order to excite electrons on higher energy levels. Then, as the electrons dropped down to lower energy levels, photons were emitted by the atoms. What is the energy of a photon emitted by a electron dropping from level n

2.56 eV

Explanation:

Using the formula;

ΔE= -RH(1/n^2final - 1/n^2initial)

RH = -2.18 * 10-18 J

nfinal = 2

ninitial =1

Substituting values;

ΔE= -2.18 * 10-18 J(1/4^2 - 1/2^2)

ΔE= -2.18 * 10-18 J(0.0625 - 0.25)

ΔE= 4.09 * 10^-19 J

To convert to eV

4.09 * 10^-19 J/1.6 * 10^-19

=2.56 eV

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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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.