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disa [49]
1 year ago
7

In his explanation of the threshold frequency in the photoelectric effect, Einstein reasoned that the absorbed photon must have

a minimum energy to dislodge an electron from the metal surface. This energy is called the work function (Ф) of the metal. What is the longest wavelength of radiation (in nm) that could cause the photoelectric effect in each of these metals:
(a) calcium, Ф=4.60 × 10⁻¹⁹J
Chemistry
1 answer:
astraxan [27]1 year ago
7 0

The wavelength of radiation that could cause the photoelectric effect in calcium is &\lambda=4.32×10^{-7} m

<h3>What is the photoelectric effect?</h3>

The photoelectric effect is a phenomenon that results in electrically charged particles being discharged from or within a substance when it absorbs electromagnetic radiation. When light strikes a metal plate, the action is frequently described as the ejection of electrons from the plate.

For calcium, the wavelength will be

$$\begin{aligned}&\mathrm{E}=4.6^{*} 10^{-19} \mathrm{~J} \\&\mathrm{~h}=6.626^{*} 10^{-34} \mathrm{Js} \\&\mathrm{c}=3^{*} 10^{8} \mathrm{~m} / \mathrm{s} \\&\mathrm{E}=\frac{h c}{\lambda} \\&\lambda=\frac{h c}{E} \\&\lambda=\frac{6.626 * 10^{-34} \mathrm{Js} * 3 * 10^{8} \mathrm{~m} / \mathrm{s}}{4.6 * 10^{-19} \mathrm{~J}} \\&\lambda=4.32^{*} 10^{-7} \mathrm{~m}\end{aligned}$$

The wavelength of radiation that could cause the photoelectric effect in calcium is &\lambda=4.32 ×10^{-7} \mathrm{~m}.

To know more about the photoelectric effect, visit: brainly.com/question/1279581

#SPJ4

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One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will
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Explanation:

The given data is as follows.

Boiling point of water (T^{o}_{b}) = 100^{o}C = (100 + 273) K = 323 K,

Boiling point of solution (T_{b}) = 101.24^{o}C = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

              \Delta T_{b} = (T_{b} - T^{o}_{b})

                           = 374.24 K - 323 K

                           = 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

            Molality = \frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}

Let molar mass of the solute is x grams.

Therefore,   Molality = \frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}

                        m = \frac{288 g \times 1000}{x g \times 90}              

                          = \frac{3200}{x}

As,    \Delta T_{b} = k_{b} \times molality

                 1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}

                       x = \frac{0.512 ^{o}C/m \times 3200}{1.24}

                          = 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is C_{n}H_{2n}O_{n}.

As, its empirical formula is CH_{2}O and mass is 30 g/mol. Hence, calculate the value of n as follows.

                n = \frac{\text{Molecular mass}}{\text{Empirical mass}}

                   = \frac{1321.29 g}{30 g/mol}

                   = 44 mol

Thus, we can conclude that the formula of given material is C_{44}H_{88}O_{44}.

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