The mass numbers of two elements X and Z are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For element $X$: Mass number $A = 52$. Let the number of protons be $p$. Number of neutrons $n = p + 0.166p = 1.166p$. Since $A = p + n$,we have $

Vedclass · Accepted Answer

${ }_{24} Cr, { }_{33} As$

Vedclass · Answer

(A) For element $X$: Mass number $A = 52$. Let the number of protons be $p$. Number of neutrons $n = p + 0.166p = 1.166p$. Since $A = p + n$,we have $52 = p + 1.166p = 2.166p$. Thus,$p = 52 / 2.166 \approx 24$. The element with atomic number $24$ is $Cr$.For element $Z$: Mass number $A = 75$. Let the number of protons be $p'$. Number of neutrons $n' = p' + 0.273p' = 1.273p'$. Since $A = p' + n'$,we have $75 = p' + 1.273p' = 2.273p'$. Thus,$p' = 75 / 2.273 \approx 33$. The element with atomic number $33$ is $As$.

The mass numbers of two elements $X$ and $Z$ are $52$ and $75$ respectively. $X$ contains $16.6 \%$ more neutrons compared to protons. $Z$ contains $27.3 \%$ more neutrons compared to protons. $X$ and $Z$ are respectively

Similar Questions

The number of protons,neutrons and electrons in ${}^{13}_{6}C$ respectively are

The number of electrons in a neutral atom of an element is equal to its

Atomic number means

Among the following, the isoelectronic species is/are:
$(I)$ $O^{2-}, F^{-}, Na^{+}, Mg^{2+}$
$(II)$ $Na^{+}, Mg^{+}, Al^{3+}, F^{-}$
$(III)$ $N^{3-}, O^{2-}, F^{-}, Ne$

The total number of isoelectronic species from the given set is $.......$
$O^{2-}, F^{-}, Al, Mg^{2+}, Na^{+}, O^{+}, Mg, Al^{3+}, F$

Vedclass Test Series

Exam Paper Generator

Online Exam Module