The energy needed to heat a 235 mL cup of water is 59.99 kJ. Using your microwav

Question

The energy needed to heat a 235 mL cup of water is 59.99 kJ. Using your microwave oven, how many moles of photons are needed to heat a 235 mL cup of water? Most microwave ovens emit microwaves that are absorbed by water at a wavelength of 12.24 cm.

At what velocity, in km, would an electron have to be moving to eject an electron from an atom of neon, which has a first ionization energy equal to 2080 kJ mol–1?

(the mass of an electron is 9.10938291 x 10–31 kg )

Calculate, to four significant figures, the wavelength in nanometers of the spectral line in Balmer series for which n2 = 4.

Explanation / Answer

1) E = hc/wavelength
E = 6.626x 10-34 * 3 x 108/12.2x 10-2 for one photon.

E = 1.63 x 10-24 J/photon

To get an energy of 59.99 kJ we will need

59.99 x 103 J/1.63 x 10-24 J = 36.7 x 1027 photons

2)

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation. It is quantitatively expressed in symbols as:

X + energy X+ + e

E = (1/2) mv2

2080 kJ mol–1 is for 6.023 x 1023 atoms so for 1 atom the energy would be

2080 x 103 J/6.023 x 1023 = 3.45 x 10-18 J

3.45 x 10-18 J = 1/2 x9.10938291 x 10–31 kg x v2

v2 = 3.45 x 10-18 x 2 /9.10938291 x 10–31

v = 2.75 x 106 m/s

v = 2.75 x 103 km/s should be the velocity of the electron

3) Use the Balmer series, which is Y = B(n2 / (n2 - 4))

where Y is the wavelength of the spectral line,

B is a constant value of approximately 364.5068 nm, and n is an integer greater than 2.

So n2= 4 cannot be calculated

I will calculate for n=4

Y = 364.5068 * 16/16-4

Y = 486.0090 nm

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