As the electron in the Bohr Orbit of hydrogen atom passes from state $n = 2$ to $n = 1$ , the $K E (K)$ and $P E (U)$ change as

K

two-fold,

U

also two-fold

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K

four-fold,

U

also four-fold

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K

four-fold,

U

two-fold

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K

two-fold,

U

four-fold

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $K$ two-fold, $U$ four-fold
As energy level at state n
$P E = 13.6 (\frac{1}{n^{2}})$
as $n = 2 \to n = 1$
$P E (n = 1) = 13.6 e V$
$P E (n = 2) = 13.6 / 4 e V$
Thus $P E$ reduced four fold times.
$- \frac{P E}{2} = K E$
Hence $K E$ will be reduced by two fold times.

Suggest Corrections

Similar questions

Q. What was Tilaks four-fold programme to mobilise masses?

Q. In paper folding method of constructing perpendicular, apart from the fold along a given line, if two more folds are made in the instructed fashion to obtain perpendicular, how many perpendiculars will be formed in total ?

What is 4-fold axis and 2-fold axis?

In women, _____(i)_______ works as a protection pad during intercourse. The fatty tissue of the protective pad also continues between the legs to surround the vaginal orifice, the inner folds are small, hairless and known as ____(ii)______. The inner folds are hidden under the two larger folds known as ___(iii)______ which are covered with hair. The entirety of all the above parts is collectively known as the ____(iv)______.

A circular sheet of paper is folded along the lines in the directions shown. The paper, after being punched in the final folded state as shown and unfolded in the reverse order of folding, will look like

Join BYJU'S Learning Program

As the electron in the Bohr Orbit of hydrogen atom passes from state n=2 to n=1, the KE(K) and PE(U) change as

The correct option is D K two-fold, U four-foldAs energy level at state nPE=13.6(1n2)as n=2→n=1PE(n=1)=13.6eVPE(n=2)=13.6/4eVThus PE reduced four fold times.−PE2=KEHence KE will be reduced by two fold times.

As the electron in the Bohr Orbit of hydrogen atom passes from state $n = 2$ to $n = 1$ , the $K E (K)$ and $P E (U)$ change as

The correct option is D $K$ two-fold, $U$ four-fold
As energy level at state n
$P E = 13.6 (\frac{1}{n^{2}})$
as $n = 2 \to n = 1$
$P E (n = 1) = 13.6 e V$
$P E (n = 2) = 13.6 / 4 e V$
Thus $P E$ reduced four fold times.
$- \frac{P E}{2} = K E$
Hence $K E$ will be reduced by two fold times.