E = hc/lambda
E = 6.624 x 10^-34 x 3 x 10^8 / 5.90 x 10^-14
E = 3.36 x 10^ 14
The correct answer is (c) It would have lower ionization energies than K and a lower electronegativity value than K
Explanation- The unknown element belongs to the alkali metals of periodic table. The alkali metal atoms have the largest size in a particular period of the period of periodic table. With the increase in the atomic number, the atom becomes larger.
Atomic radius decreases --> ionization energy increase---->Electronegativity increases. We can use this flow chart to compare the IE and EN of the element. As K have lower radii than the unknown element, so the IE and EN of K will be greater than the unknown element.
Answer: Artificial transmutation may occur in machinery that has enough energy to cause changes in the nuclear structure of the elements. This releases, on average, 3 neutrons and a large amount of energy. The released neutrons then cause fission of other uranium atoms, until all of the available uranium is exhausted.
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3. A
4. B
5. A
6. E
7. A
8. C
9. A
10. B
Some of these were guesses but they were educated guesses. Best of luck. If some of them are wrong I am sorry. <span />