An element has an atomic weight of 40 and its | vedclass

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(A) The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ corresponds to a total of $2 + 2 + 6 + 2 + 6 = 18$ electrons.For a neutral atom,the atomic

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$18$ and $22$

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(A) The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ corresponds to a total of $2 + 2 + 6 + 2 + 6 = 18$ electrons.For a neutral atom,the atomic number $(Z)$ is equal to the number of electrons,so $Z = 18$.The atomic mass $(A)$ is given as $40$.The number of neutrons $(n)$ is calculated using the formula: $n = A - Z$.$n = 40 - 18 = 22$.Therefore,the atomic number is $18$ and the number of neutrons is $22$.

An element has an atomic weight of $40$ and its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. What are its atomic number and the number of neutrons?

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