Question

Few electrons have following quantum numbers,
$\left(i\right)n=4,l=1\left(ii\right)n=4,l=0$
$\left(iii\right)n=3,l=2\left(iv\right)n=3,l=1$
Arrange them in the order of increasing energy from lowest to highest.

• A. $\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$
• B. $\left(ii\right)<\left(iv\right)<\left(i\right)<\left(iii\right)$
• C. $\left(i\right)<\left(iii\right)<\left(ii\right)<\left(iv\right)$
• D. $\left(iii\right)<\left(i\right)<\left(iv\right)<\left(ii\right)$

Answer 1

Answer: (A)

$\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$

$\left(i\right)n=4,l=1\to 4p$

$\left(ii\right)n=4,l=0\to 4s$

$\left(iii\right)n=3,l=2\to 3d$

$\left(iv\right)n=3,l=1\to 3p$

Energy order

$1s<2s<2p<3s<3p<4s<3d<4p$

$\Rightarrow \left(iv\right)<\left(ii\right)<\left(iii\right)<\left(1\right)$

also could be calculated or basis of $n+1$ order value