Chapter 27, Problem 28P

To determine

(a) To determine:

The value of Planck’s constant

Answer to Problem 28P

Solution:

The value of Planck’s constant is $6.53\times {10}^{-34}\text{J-sec}$.

Explanation of Solution

Given:

Table of incident frequency and energy of ejected electron

Frequency(1014 Hz)	Energy (eV)
11.8	2.60
10.6	2.11
9.9	1.81
9.1	1.47
8.2	1.10
6.9	0.57

Calculation:

The graph between Energy and frequency is given as follows

Now using the first value the equation of energy of electron is given as

$h\times 11.8\times {10}^{14}=W_o+2.60\times 1.6\times {10}^{-19}$ …… (1)

Now using the second value the equation of energy of electron is given as

$h\times 10.6\times {10}^{14}=W_o+2.11\times 1.6\times {10}^{-19}$ …… (2)

Subtracting Eq. (1)with Eq. (2)

$\begin{array}{c}h=\frac{\left(2.60-2.11\right)\times 1.6\times {10}^{-19}}{\left(11.8-10.6\right)\times {10}^{14}}\\ =6.53\times {10}^{-34}\text{J-sec}\end{array}$

To determine

(b) To determine:

The cutoff frequency of the sodium.

Answer to Problem 28P

Solution:

The cutoff frequency of the sodium $5.42\times {10}^{14}\text{Hz}$.

Explanation of Solution

The cut off frequency is the frequency at which the energy of the electron is 0

Consider the Eq. (1)

$h\times 11.8\times {10}^{14}=W_o+2.60\times 1.6\times {10}^{-19}$

Plugging the value of Planck’s constant

$\begin{array}{c}6.53\times {10}^{-34}\times 11.8\times {10}^{14}=W_o+2.60\times 1.6\times {10}^{-19}\\ W_o=3.54\times {10}^{-19}\text{J}\end{array}$

Now the cutoff frequency is calculated as

$W_o=hf$

Plugging the values in the above equation

$\begin{array}{c}3.54\times {10}^{-19}=\text{6}\text{.53}\times {\text{10}}^{-34}\times f\\ f=5.42\times {10}^{14}\text{Hz}\end{array}$

Thus, the cutoff frequency is $5.42\times {10}^{14}\text{Hz}$.

To determine

(c) To determine:

Work function of the sodium

Answer to Problem 28P

Solution:

Work function of the sodium is $3.54\times {10}^{-19}\text{J}$.

Explanation of Solution

Consider the Eq. (1)

$h\times 11.8\times {10}^{14}=W_o+2.60\times 1.6\times {10}^{-19}$

Plugging the value of Planck’s constant

$\begin{array}{c}6.53\times {10}^{-34}\times 11.8\times {10}^{14}=W_o+2.60\times 1.6\times {10}^{-19}\\ W_o=3.54\times {10}^{-19}\text{J}\end{array}$