The formation of a meander. As the river erodes laterally, to the right side then the left side, it forms large bends, and then horseshoe-like loops called meanders . The formation of meanders is due to both deposition and erosion and meanders gradually migrate downstream.

5 0

2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago

The element that has a valence configuration of 5s25p6 is ________.

Ivanshal [37]

Its period 5 from 5s25p6, with Xenon(54) as the noble gas. 2+6 = 8 electrons

54+8 = 62, or Sm.

3 0

3 years ago

Help me please!!!!!!!!!!!!!!!

Alex787 [66]

Answer:

2KBr + MgF₂ –> 2KF + MgBr₂

The coefficients are: 2, 1, 2, 1

Explanation:

KBr + MgF₂ –> KF + MgBr₂

The above equation can be balance as illustrated below:

KBr + MgF₂ –> KF + MgBr₂

There are 2 atoms of F on the left side and 1 atom on the right. It can be balance by writing 2 before KF as shown below:

KBr + MgF₂ –> 2KF + MgBr₂

There 2 atoms of K on the right side and 1 atom on the left side. It can be balance by writing 2 before KBr as shown below:

2KBr + MgF₂ –> 2KF + MgBr₂

Now, the equation is balanced.

The coefficients are: 2, 1, 2, 1

8 0

2 years ago