Answer: E
=
1.55
⋅
10
−
19
J
Explanation:
The energy transition will be equal to 1.55
⋅
10
−
1
J
.
So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition
1
λ =
R
⋅
(
1
n
2
final −
1
n
2
initial )
, where
λ
- the wavelength of the emitted photon;
R
- Rydberg's constant - 1.0974
⋅
10
7
m
−
1
;
n
final
- the final energy level - in your case equal to 3;
n
initial
- the initial energy level - in your case equal to 5.
So, you've got all you need to solve for λ
, so
1
λ =
1.0974
⋅10 7
m
−
1
⋅
(....
−152
)
1
λ
=
0.07804
⋅
10
7
m
−
1
⇒
λ
=
1.28
⋅
10
−
6
m
Since
E
=
h
c
λ
, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by
h
⋅
c
, where
h
- Planck's constant -
6.626
⋅
10
−
34
J
⋅
s
c
- the speed of light -
299,792,458 m/s
So, the transition energy for your particular transition (which is part of the Paschen Series) is
E
=
6.626
⋅
10
−
34
J
⋅
s
⋅
299,792,458
m/s
1.28
⋅
10
−
6
m
E
=
1.55
⋅
10
−
19
J
Answer:
Hydrogen is the chemical element with the symbol H and atomic number 1. With a standard atomic weight of 1.008, hydrogen is the lightest element in the periodic table.
Explanation:
Answer:
The change in energy when a neutral atom in the gaseous state adds an electron to form a negative ion.
Explanation:
That is, it is the energy involved in the reaction
X(g) + e⁻ ⟶ X⁻(g)
For most elements, the electron affinity is negative.
However, there are two major exceptions — the values are positive for the elements of Groups 2 and 18 (note the troughs in the graph below).
Answer: it they are both in the same place
Explanation:I don’t know and don’t care loser
Answer:
<u>ATGGCCTA</u>
Explanation:
For this we have to keep in mind that we have a <u>specific relationship between the nitrogen bases</u>:
-) <u>When we have a T (thymine) we will have a bond with A (adenine) and viceversa</u>.
-) <u>When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa</u>.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
<u>ATGGCCTA</u>
<u></u>
I hope it helps!
