Question

Neon gas is generally used in signboards. If it emits strongly at \[616nm\].
Calculate:
a) The frequency of emission.
b) Distance travelled by this radiation in \[30s\]
c) The energy of quantum and
d) The number of quanta present, if it produces \[2J\] of energy.

Answer 1

Hint: The frequency and wavelength are inversely proportional to each other, and the frequency can be expressed in terms of wavelength and velocity of light. The velocity can be expressed in terms of distance and time. The energy is expressed in Planck’s constant and frequency.
Formula used:
\[\upsilon = \frac{c}{\lambda }\]
\[\upsilon \] is frequency of emission
\[\lambda \] is wavelength
\[c\] is velocity of light which is constant \[3 \times {10^8}m\]
\[c = \frac{d}{t}\]
\[t\] is time in seconds
\[d\] is distance
\[E = h\upsilon \]
\[E\] is energy
\[h\] is Planck’s constant \[ = 6.626 \times {10^{ - 34}}\]

Complete answer:
A.The wavelength given is \[616nm\] But one nanometre will be equal to \[{10^{ - 9}}m\]
Substitute the values in the formula, will get
\[\upsilon = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{616 \times {{10}^{ - 9}}}} = 4.87 \times {10^{14}}{\sec ^{ - 1}}\]
Thus, the frequency is \[4.87 \times {10^{14}}{\sec ^{ - 1}}\]
B.The distance travelled in \[30s\] will be calculated by substituting the above values in the formula
\[d = t \times \lambda \times \upsilon = 30 \times 616 \times {10^{ - 9}} \times 4.87 \times {10^{14}} = 9 \times {10^9}m\]
The distance travelled in \[30s\] is \[9 \times {10^9}m\]
C.The energy of quantum can be determined from Planck’s constant and frequency
Planck’s constant \[ = 6.626 \times {10^{ - 34}}\]
The frequency is \[4.87 \times {10^{14}}{\sec ^{ - 1}}\]
The energy of quantum is
\[E = h\upsilon = 6.626 \times {10^{ - 34}} \times 4.87 \times {10^{14}} = 3.27 \times {10^{ - 19}}J\]
Thus, energy of quantum is \[3.27 \times {10^{ - 19}}J\]
The energy produced is \[2J\]
Substitute this energy in the value of energy
D.The number of quanta present is
\[\frac{2}{{3.27 \times {{10}^{ - 19}}}} = 6.2 \times {10^{18}}\]photons
Thus, the number of quanta present is \[6.2 \times {10^{18}}\] photons, if \[2J\] of energy is produced.

Note:
The units must be in meters while calculating the frequency as wavelength given is in nanometres. The conversion must be correctly done. The Planck’s constant has also taken in joules seconds. Energy must be in joules and quanta is described as a set of photons and thus written as photons.