Answer:
4.59 × 10⁻³⁶ kJ/photon
Explanation:
Step 1: Given and required data
- Wavelength of the violet light (λ): 433 nm
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
- Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Convert "λ" to meters
We will use the conversion factor 1 m = 10⁹ nm.
433 nm × 1 m/10⁹ nm = 4.33 × 10⁷ m
Step 3: Calculate the energy (E) of the photon
We will use the Planck-Einstein's relation.
E = h × c/λ
E = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/4.33 × 10⁷ m
E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ
Answer:
The atomic number of burienium will be 307.
Explanation:
During positron emission proton is converted into the neutron and one electron neutrino with positron is released. It means the atomic number will be reduce by one and atomic mass remain same.
For example:
²³Mg₁₂ → ₁₁Na²³+ e⁺+ Ve
Similarly, when highlinium-308 undergoes positron emission the new element burienium is produced and the atomic number will be 307 while atomic mass remain same.
Properties of beta radiations:
Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.
The mass of beta particle is smaller than the alpha particles.
They can travel in air in few meter distance.
These radiations can penetrate into the human skin.
The sheet of aluminium is used to block the beta radiation
