Hi,
The atomic number of an element depends on how many protons there are in one atom of that element. If the atom has 3 protons, its atomic number will be 3.
Hope this helps! If my answer was not clear enough or you’d like further explanation please let me know.
<u>Answer:</u> The potential of electrode is -0.79 V
<u>Explanation:</u>
When zinc is dipped in zinc sulfate solution, the electrode formed is 
Reduction reaction follows: 
To calculate the potential of electrode, we use the equation given by Nernst equation:
![E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}](https://tex.z-dn.net/?f=E_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D%3DE%5Eo_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5D%7D%7B%5BZn%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = -0.76 V
n = number of electrons exchanged = 2
![[Zn]=1M](https://tex.z-dn.net/?f=%5BZn%5D%3D1M)
(concentration of pure solids are taken as 1)
![[Zn^{2+}]=0.1M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.1M)
Putting values in above equation, we get:
Hence, the potential of electrode is -0.79 V
Answer:
table C will be the correct answer
<h2>Ultraviolet Light</h2>
Explanation:
- The energy of a photon that will be released if an electron falls from the n= 2 orbit (excited state) to the n0 = 1 orbit (ground state) is of ultraviolet light.
- In the ultraviolet part of the spectrum, a photon having an energy of 10.2 eV has a wavelength of 1.21 x 10-7 m.
- Hence, when an electron wants to jump or it gets excited from the first level to the second level that is from n = 1 orbit to n = 2 orbits, it must absorb a photon of ultraviolet light.
- But,When an electron falls from n = 2 orbit to n = 1 orbit or from n = 2 orbit(excited state) to n = 0 orbit(groubd state), it emits a photon of ultraviolet light.
