Question

N=1 is the lowest electronic energy state for a hydrogen atom. (a) If a hydrogen atom is in a state N=4, what is K+U for this atom (in eV)? (b) The hydrogen atom makes a transition to state N=2, Now what is K+U in electron volts for this atom? (c) What is energy (in eV) of the photon emitted in the transition from level N=4 to N=2? (d) Which of the arrows in figure 8.40 represents this transition?

Short answer

The total energy level in the fourth level is $-0.85eV$.

Step-by-step solution

Identification of given data

The state of hydrogen atom is $N=4$

Conceptual Explanation

The energy of an atom at a particular level varies with the level of the atom. The energy in the lowest level of hydrogen atom is 13.6 eV.

Determination of total energy of hydrogen atom

The total energy of hydrogen atom is given as:

${\left(K+U\right)}_N=-\frac{13.6\text{eV}}{N^2}$

Substitute all the values in the above equation.

$$\begin{gathered} {\left(K+U\right)}_4=-\frac{13.6\text{eV}}{{\left(4\right)}^2} \\ {\left(K+U\right)}_4=-0.85\text{eV} \end{gathered}$$

Therefore, the total energy of hydrogen is $-0.85eV$