The frequency V of the radiation emitted by an atom when an electron jumps from one orbit to another is given by v=k δ E, where k is a constant and δ E is the change in energy level due to the transition. Then dimension of k is

Question

The frequency V of the radiation emitted by an atom when an electron jumps from one orbit to another is given by v=k δ E, where k is a constant and δ E is the change in energy level due to the transition. Then dimension of k is (A) M L2 T-2 (B) the same dimension of angular momentum (C) M L2 T-1 (D) M-1 L-2 T

Tardigrade Admin · Accepted Answer

v=k δ E [k]=[(v/δ E)]=([T-1]/[M L2 T-2]) [k]=[M-1 L-2 T]

Q. The frequency $V$ of the radiation emitted by an atom when an electron jumps from one orbit to another is given by $v = k δ E$ , where $k$ is a constant and $δ E$ is the change in energy level due to the transition. Then dimension of $k$ is

$M L^{2} T^{- 2}$

the same dimension of angular momentum

$M L^{2} T^{- 1}$

$M^{- 1} L^{- 2} T$

$v = k δ E$
$[k] = [\frac{v}{δ E}] = \frac{[ T ^{- 1} ]}{[ M L ^{2} T ^{- 2} ]}$
$[k] = [M^{- 1} L^{- 2} T]$

Q. The frequency V of the radiation emitted by an atom when an electron jumps from one orbit to another is given by v=kδE, where k is a constant and δE is the change in energy level due to the transition. Then dimension of k is

ML2T−2

the same dimension of angular momentum

ML2T−1

M−1L−2T

v=kδE [k]=[δEv​]=[ML2T−2][T−1]​ [k]=[M−1L−2T]

Q. The frequency $V$ of the radiation emitted by an atom when an electron jumps from one orbit to another is given by $v = k δ E$ , where $k$ is a constant and $δ E$ is the change in energy level due to the transition. Then dimension of $k$ is

$M L^{2} T^{- 2}$

$M L^{2} T^{- 1}$

$M^{- 1} L^{- 2} T$

$v = k δ E$
$[k] = [\frac{v}{δ E}] = \frac{[ T ^{- 1} ]}{[ M L ^{2} T ^{- 2} ]}$
$[k] = [M^{- 1} L^{- 2} T]$