An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is (k h R/25 m). Find k. (Here, m= Mass of electron, h= Planck's constant and R= Rydberg constant)

Question

An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is (k h R/25 m). Find k. (Here, m= Mass of electron, h= Planck's constant and R= Rydberg constant) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

p=m v (h/λ)=m v; ∵ (1/λ)=R((1/n12)-(1/n22)) ⇒ m v=h R[(1/n12)-(1/n22)] v=(h R/m)[(1/12)-(1/52)]=(24 h R/25 m)

Q. An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is $\frac{kh R}{25 m}$ . Find $k$ .
(Here, $m =$ Mass of electron, $h =$ Planck's constant and $R =$ Rydberg constant)

$p = m v$
$\frac{h}{λ} = m v$ ;
$∵ \frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$\Rightarrow m v = h R [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$
$v = \frac{h R}{m} [\frac{1}{1 ^{2}} - \frac{1}{5 ^{2}}] = \frac{24 h R}{25 m}$

Q. An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is 25mkhR​. Find k. (Here, m= Mass of electron, h= Planck's constant and R= Rydberg constant)

p=mv λh​=mv; ∵λ1​=R(n12​1​−n22​1​) ⇒mv=hR[n12​1​−n22​1​] v=mhR​[121​−521​]=25m24hR​

Q. An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is $\frac{kh R}{25 m}$ . Find $k$ .
(Here, $m =$ Mass of electron, $h =$ Planck's constant and $R =$ Rydberg constant)

$p = m v$
$\frac{h}{λ} = m v$ ;
$∵ \frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$\Rightarrow m v = h R [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$
$v = \frac{h R}{m} [\frac{1}{1 ^{2}} - \frac{1}{5 ^{2}}] = \frac{24 h R}{25 m}$