Based on the equation Delta E = - 2.0 times 1 | vedclass

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

Question

(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{2^2} -

Vedclass · Accepted Answer

$1.325 	imes 10^{-7} \, m$

Vedclass · Answer

(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{2^2} - \frac{1}{1^2} ight)$$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{4} - 1 ight) = - 2.0 	imes 10^{-18} 	imes \left( - \frac{3}{4} ight) = 1.5 	imes 10^{-18} \, J$Using the relation $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is:$\lambda = \frac{hc}{\Delta E} = \frac{6.625 	imes 10^{-34} 	imes 3 	imes 10^8}{1.5 	imes 10^{-18}}$$\lambda = 1.325 	imes 10^{-7} \, m$

Similar Questions

What is the maximum number of different spectral lines obtained when a sample of a large number of $H$-atoms in the $5^{th}$ excited state makes transitions to the ground state,given that no lines are emitted in the Balmer series?

The most probable radius (in $pm$) for finding the electron in $He^{+}$ is

Which of the following provisions of the Bohr atomic model of hydrogen turned out to be false?

If the electron of a hydrogen atom is present in the first orbit,the total energy of the electron is

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to excite the electron in a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is (Note: The question asks for energy per atom in Joules. Given $1 \ eV = 1.602 \times 10^{-19} \ J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module