The ionization energy for potassium 419 kj/mol. the wavelength of light emitted when an excited k atom undergoes the 4s< 4p transition is approximately 769 nm. using this info, calculate the energies of the 4s and 4p orbitals in potassium.

Answer : The energies of the 4s and 4p orbitals in potassium is,

Solution :

First we have to calculate the change in energy.

Formula used :

where,

= change in energy

h = Planck's constant =

c = speed of light =

=

Now put all the given values in this formula, we get

The difference in energy levels is defined as,

Now we have to calculate the first ionization energy for one atom of potassium.

Now this ionization energy will be equal to the negative of the orbital energy of the electron located in the 4s-orbital.

Now we have to calculate the energy of the 4p orbital in potassium.

Therefore, the energies of the 4s and 4p orbitals in potassium is,

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol