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Question: The photoelectric effect is applied in some burglar alarms. A beam of light shines on a metal electr...

The photoelectric effect is applied in some burglar alarms. A beam of light shines on a metal electrode in a vacuum tube(photoelectric cell) and causes electrons to be photo electrically ejected from the surface of the metal. This electrode thus becomes the cathode. The ejected electrons are attracted to the anode in the vacuum tube and the electric circuit is completed using a battery. If the light beam is blocked by a burglar’s alarm, the electric circuit is broken. This sets off the alarm system. What is the maximum wavelength of light that you could use for a burglar alarm if the cathode of the photoelectric cell is made of tungsten and electrons are ejected from tungsten with a kinetic energy of 8.0×10−128.0 \times {10^{ - 12}} erg when the wavelength of the incident light is exactly 1.25×103 Ao?1.25 \times {10^3}\mathop {{\text{ }}A}\limits^o ?
A. 2.5×103A∘2.5 \times {10^3}{A^ \circ }
B. 3.4×103A∘3.4 \times {10^3}{A^ \circ }
C. 4.5×103A∘4.5 \times {10^3}{A^ \circ }
D. 5.6×103A∘5.6 \times {10^3}{A^ \circ }

Explanation

Solution

Photoelectric effect is defined as the phenomena of the ejection of an electron from the surface of the metal when the light of suitable frequency falls on it. This phenomenon helps us understand the quantum nature of light and electrons.

Complete step by step answer:
Given the kinetic energy of an ejected electron from the tungsten = 8.0×10−128.0 \times {10^{ - 12}}
erg and the wavelength of the incident light is 1.25×103 Ao1.25 \times {10^3}\mathop {{\text{ }}A}\limits^o
or 1.25×103×10−8cm1.25 \times {10^3} \times {10^{ - 8}}cm. As we know the formula for the kinetic energy of the ejected electron is KE=hν−hv∘KE = h\nu - h{v_ \circ }
Here h = planck's constant which has value 6.625×10−27ergsec⁡.6.625 \times {10^{ - 27}}erg\sec .
And ν\nu is the frequency and ν∘{\nu _\circ } is threshold frequency.
As we know the relation between frequency and wavelength is given as ν=cλ\nu = \dfrac{c}{\lambda }
Here ccis the velocity of light and its value is 3×1010cm/sec⁡3 \times {10^{10}}cm/\sec . Hence we can write:
KE=hcλ−hcλoKE = h\dfrac{c}{\lambda } - h\dfrac{c}{{{\lambda _o}}}
⇒KE=hc(1λ−1λo)\Rightarrow KE = hc(\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}})
On putting the value in the above equation we get:
8×10−12=6.625×10−27×3×1010×(11.25×103×10−8−1λo)8 \times {10^{ - 12}} = 6.625 \times {10^{ - 27}} \times 3 \times {10^{10}} \times (\dfrac{1}{{1.25 \times {{10}^3} \times {{10}^{ - 8}}}} - \dfrac{1}{{{\lambda _o}}})

⇒8×10−126.625×10−27×3×1010=11.25×103×10−8−1λo \Rightarrow \dfrac{{8 \times {{10}^{ - 12}}}}{{6.625 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}}}} = \dfrac{1}{{1.25 \times {{10}^3} \times {{10}^{ - 8}}}} - \dfrac{1}{{{\lambda _o}}}
⇒1λo=11.25×103×10−8−8×10−126.625×10−27×3×1010\Rightarrow \dfrac{1}{{{\lambda _o}}} = \dfrac{1}{{1.25 \times {{10}^3} \times {{10}^{ - 8}}}} - \dfrac{{8 \times {{10}^{ - 12}}}}{{6.625 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}}}}
On solving we get:
λo=2.5×10−5cm{\lambda _o} = 2.5 \times {10^{ - 5}}cm
As we know 1 AoA^{o} = 10−810^{-8}. Hence λo=2.5×10−510−8=2.5×10−5+8{\lambda _o} = \dfrac{{2.5 \times {{10}^{ - 5}}}}{{{{10}^{ - 8}}}} = 2.5 \times {10^{ - 5 + 8}}
Hence the correct answer is option A.

Additional Information: Threshold energy is not same for each metal and differs from metal to metal and hence the green light might cause photoelectric effect for one metal and may be ineffective for other metal in causing the photoelectric effect

Note: The color of light emitted during photoelectric effect depends on the frequency of light. For example, the frequency of red light is less than that of blue light. There are two concepts related to photoelectric effect, i.e. threshold energy and threshold frequency. Here, threshold energy is defined as the minimum amount of energy that is required to remove an electron and threshold frequency is the minimum frequency of light so that photoelectric effect takes place.