(Take Planck's constant as $6.6 \times 10^{-34} \text{ J s}$)
Understanding the Photoelectric Effect Condition
The photoelectric effect happens only if the energy of the incident photon ($E_{photon}$) meets or exceeds the metal's work function ($\phi$).
Photon energy is calculated using $E_{photon} = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
The condition for the effect is: $ \frac{hc}{\lambda} \ge \phi $ This means the effect occurs for wavelengths $\lambda \le \frac{hc}{\phi}$.
Conversely, the effect *does not* occur if the photon energy is less than the work function: $ \frac{hc}{\lambda} < \phi $ This condition implies the effect is absent for wavelengths $\lambda > \frac{hc}{\phi}$. The critical wavelength $\lambda_0 = \frac{hc}{\phi}$ is known as the threshold wavelength.
Calculating the Threshold Wavelength
We are given:
- Work function, $\phi = 6.6 \text{ eV}$
- Planck's constant, $h = 6.6 \times 10^{-34} \text{ J s}$
- Speed of light, $c \approx 3 \times 10^8 \text{ m/s}$
- Conversion factor, $1 \text{ eV} \approx 1.602 \times 10^{-19} \text{ J}$
First, calculate the value of $hc$:
$ hc = (6.6 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s}) = 1.98 \times 10^{-25} \text{ J m} $Next, convert the work function $\phi$ from electron volts (eV) to Joules (J):
$ \phi = 6.6 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) \approx 1.05732 \times 10^{-18} \text{ J} $Now, calculate the threshold wavelength $\lambda_0$ using the formula $\lambda_0 = \frac{hc}{\phi}$:
$ \lambda_0 = \frac{1.98 \times 10^{-25} \text{ J m}}{1.05732 \times 10^{-18} \text{ J}} \approx 1.8726 \times 10^{-7} \text{ m} $Convert this threshold wavelength to nanometers (nm):
$ \lambda_0 \approx 1.8726 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} \approx 187.3 \text{ nm} $Identifying the Wavelength Without Photoelectric Effect
The photoelectric effect will not occur for incident radiation wavelengths ($\lambda$) that are greater than the threshold wavelength ($\lambda_0$).
We need to find the option where $\lambda > 187.3 \text{ nm}$. Let's check the given options:
- Option 1: $50 \text{ nm}$ ($50 < 187.3$) - Effect occurs.
- Option 2: $100 \text{ nm}$ ($100 < 187.3$) - Effect occurs.
- Option 3: $150 \text{ nm}$ ($150 < 187.3$) - Effect occurs.
- Option 4: $200 \text{ nm}$ ($200 > 187.3$) - Effect does NOT occur.
Therefore, the wavelength of incident radiation that does not give rise to the photoelectric effect is $200 \text{ nm}$.
Similar Questions
In the first excited state of hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately :
(Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}, \text{ e} = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)Four statements are given (A is mass number) :
A. The volume of a nucleus is proportional to $A^{1/3}$.
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituent nucleons is called the mass defect.
Choose the correct answer from the options given below :- An unknown nucleus has a nuclear density of $2.29 \times 10^{17} \text{ kg/m}^3$ and mass of $19.926 \times 10^{-27} \text{ kg}$. Its mass number A is approximately :
(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$) - In Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot ?
- A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by :
- A ray of light with wavelength $\lambda$ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 \gg \lambda$. The correct option is :
An ideal Zener diode with breakdown voltage of $-3\text{ V}$ is reverse biased with a negative input voltage $V_i = -5\text{ V}$. The magnitude of voltage difference between points B and A is :
- Consider the following nuclear reaction :
$$^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He}$$
Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^{4}\text{He}$ as $238.050\text{ u}$, $234.043\text{ u}$ and $4.003\text{ u}$, respectively. The Q value for the reaction, in keV, is :
[Given : $1\text{ u} = 931.5\text{ MeV c}^{-2}$] - A photon and an electron, each of $20\text{ eV}$ energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{ph}$, $\frac{p_e}{p_{ph}}$ is :
(Take speed of light $= 3 \times 10^8\text{ ms}^{-1}$, charge of electron $= -1.6 \times 10^{-19}\text{ C}$ and mass of electron $= 9 \times 10^{-31}\text{ kg}$) - Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\text{ ms}^{-1}$. The radius of the orbit is $x \times 10^{-9}\text{ m}$. The value of $x$ is :
[Take the mass of electron to be $9 \times 10^{-31}\text{ kg}$, charge of electron $= 1.6 \times 10^{-19}\text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$]
Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
