Answer:
E = 3.77×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of absorption line = 527 nm (527×10⁻⁹m)
Energy of absorption line = ?
Solution:
Formula:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = speed of wave = 3×10⁸ m/s
by putting values,
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m
E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m
E = 0.0377×10⁻¹⁷ J
E = 3.77×10⁻¹⁹ J
When an organism is buried quickly there is less decay and better the chance for it to be persevere. The hard parts of the organism such as bones, shells, and teeth have a better chance of becoming fossils that softer parts of the organism. HARD BONES.
In a beta emission, the mass number of the daughter nucleus remains unchanged while the atomic number of the daughter nucleus increases by one unit. The following are isotopes produced when the following undergo beta emission;
1) potassium-42 ------> Ca - 42
2) iodine-131 ------------> Xe - 131
3) iron-52 ---------------> Co - 52
4) sodium-24 -----------> Mg -24
The daughter nucleus formed after beta emission is found one place after its parent in the periodic table.
Regarding the stability of the daughter nuclei, a nucleus is unstable if the neutron-proton ratio is less than 1 or greater than 1.5.
Hence, the following daughter nuclei are stable; Ca - 42, Xe - 131, Mg -24.
Learn more: brainly.com/question/1371390
Answer:
[Ne]3s2
Explanation:
ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.
considerando la configuración electrónica más externa de cada una de las especies mostradas;
para la primera configuración, ns2 np6 corresponde a un gas noble.
para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.
para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.
para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6
The answer is neutrons and protons.
Hope you got it!!
