238720
The wavelengths of two photons are $2000 \AA$ and $4000 \AA$ respectively. What is the ratio of their energies?
1 $1 / 4$
2 4
3 $1 / 2$
4 2
Explanation:
: Given that, $\lambda_1=2000 \AA, \lambda_2=4000 \AA$ We know that, $\therefore \quad \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ So, $\quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{hc}}{\lambda_1} \times \frac{\lambda_2}{\mathrm{hc}}$ $\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{4000 \AA}{2000 \AA} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=2 \end{aligned}$
VITEEE 2019
Structure of Atom
238721
The values of Planck's constant is $6.63 \times 10^{-34}$ Js. The velocity of light is $3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8 \times 10^{15}$
238722
The ratio of slopes of $K_{\max } \mathrm{vs} v$ and $v_0$ vs $v$ curves in the photoelectric effects gives $(v=$ frequency, $K_{\max }=$ maximum kinetic energy, $v_0$ = stopping potential)
1 The ratio of Planck's constant of electronic charge
2 Work function
3 Planck's constant
4 Charge of electron
Explanation:
: We know that, $\mathrm{h} v=\mathrm{h} v_0+\mathrm{e} v_0$ $\mathrm{e} v_0=\mathrm{h} v-\mathrm{h} v_0, \quad v_0=\frac{\mathrm{h}}{\mathrm{e}} v-\frac{\mathrm{h}}{\mathrm{e}} v_0$ On comparing this equation with the straight line equation i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $v_0$ vs $v$ is $\begin{aligned} & \text { (slope) })_1=\frac{\mathrm{h}}{\mathrm{e}} \\ & \mathrm{h} v=\mathrm{h} v_0+\mathrm{KE} \\ & \mathrm{KE}=\mathrm{h} v-\mathrm{h} v_0 \text { (constant) } \\ & \mathrm{y}=\mathrm{mx}+\left(-\mathrm{h} v_0\right) \end{aligned}$ Thus, slope of $\mathrm{K}_{\max }$ vs $v$ is $(\text { slope })_2=\mathrm{h}$ $\therefore \frac{(\text { slope })_2}{(\text { slope })_1}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$ (charge of electron)
VITEEE 2015
Structure of Atom
238723
A certain metal when irradiated by light $(r=$ $3.2 \times 10^{16} \mathrm{~Hz}$ ) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $(r=2.0 \times$ $\left.10^{16} \mathrm{~Hz}\right)$. The $v_0$ of metal is
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Structure of Atom
238720
The wavelengths of two photons are $2000 \AA$ and $4000 \AA$ respectively. What is the ratio of their energies?
1 $1 / 4$
2 4
3 $1 / 2$
4 2
Explanation:
: Given that, $\lambda_1=2000 \AA, \lambda_2=4000 \AA$ We know that, $\therefore \quad \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ So, $\quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{hc}}{\lambda_1} \times \frac{\lambda_2}{\mathrm{hc}}$ $\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{4000 \AA}{2000 \AA} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=2 \end{aligned}$
VITEEE 2019
Structure of Atom
238721
The values of Planck's constant is $6.63 \times 10^{-34}$ Js. The velocity of light is $3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8 \times 10^{15}$
238722
The ratio of slopes of $K_{\max } \mathrm{vs} v$ and $v_0$ vs $v$ curves in the photoelectric effects gives $(v=$ frequency, $K_{\max }=$ maximum kinetic energy, $v_0$ = stopping potential)
1 The ratio of Planck's constant of electronic charge
2 Work function
3 Planck's constant
4 Charge of electron
Explanation:
: We know that, $\mathrm{h} v=\mathrm{h} v_0+\mathrm{e} v_0$ $\mathrm{e} v_0=\mathrm{h} v-\mathrm{h} v_0, \quad v_0=\frac{\mathrm{h}}{\mathrm{e}} v-\frac{\mathrm{h}}{\mathrm{e}} v_0$ On comparing this equation with the straight line equation i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $v_0$ vs $v$ is $\begin{aligned} & \text { (slope) })_1=\frac{\mathrm{h}}{\mathrm{e}} \\ & \mathrm{h} v=\mathrm{h} v_0+\mathrm{KE} \\ & \mathrm{KE}=\mathrm{h} v-\mathrm{h} v_0 \text { (constant) } \\ & \mathrm{y}=\mathrm{mx}+\left(-\mathrm{h} v_0\right) \end{aligned}$ Thus, slope of $\mathrm{K}_{\max }$ vs $v$ is $(\text { slope })_2=\mathrm{h}$ $\therefore \frac{(\text { slope })_2}{(\text { slope })_1}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$ (charge of electron)
VITEEE 2015
Structure of Atom
238723
A certain metal when irradiated by light $(r=$ $3.2 \times 10^{16} \mathrm{~Hz}$ ) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $(r=2.0 \times$ $\left.10^{16} \mathrm{~Hz}\right)$. The $v_0$ of metal is
238720
The wavelengths of two photons are $2000 \AA$ and $4000 \AA$ respectively. What is the ratio of their energies?
1 $1 / 4$
2 4
3 $1 / 2$
4 2
Explanation:
: Given that, $\lambda_1=2000 \AA, \lambda_2=4000 \AA$ We know that, $\therefore \quad \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ So, $\quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{hc}}{\lambda_1} \times \frac{\lambda_2}{\mathrm{hc}}$ $\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{4000 \AA}{2000 \AA} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=2 \end{aligned}$
VITEEE 2019
Structure of Atom
238721
The values of Planck's constant is $6.63 \times 10^{-34}$ Js. The velocity of light is $3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8 \times 10^{15}$
238722
The ratio of slopes of $K_{\max } \mathrm{vs} v$ and $v_0$ vs $v$ curves in the photoelectric effects gives $(v=$ frequency, $K_{\max }=$ maximum kinetic energy, $v_0$ = stopping potential)
1 The ratio of Planck's constant of electronic charge
2 Work function
3 Planck's constant
4 Charge of electron
Explanation:
: We know that, $\mathrm{h} v=\mathrm{h} v_0+\mathrm{e} v_0$ $\mathrm{e} v_0=\mathrm{h} v-\mathrm{h} v_0, \quad v_0=\frac{\mathrm{h}}{\mathrm{e}} v-\frac{\mathrm{h}}{\mathrm{e}} v_0$ On comparing this equation with the straight line equation i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $v_0$ vs $v$ is $\begin{aligned} & \text { (slope) })_1=\frac{\mathrm{h}}{\mathrm{e}} \\ & \mathrm{h} v=\mathrm{h} v_0+\mathrm{KE} \\ & \mathrm{KE}=\mathrm{h} v-\mathrm{h} v_0 \text { (constant) } \\ & \mathrm{y}=\mathrm{mx}+\left(-\mathrm{h} v_0\right) \end{aligned}$ Thus, slope of $\mathrm{K}_{\max }$ vs $v$ is $(\text { slope })_2=\mathrm{h}$ $\therefore \frac{(\text { slope })_2}{(\text { slope })_1}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$ (charge of electron)
VITEEE 2015
Structure of Atom
238723
A certain metal when irradiated by light $(r=$ $3.2 \times 10^{16} \mathrm{~Hz}$ ) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $(r=2.0 \times$ $\left.10^{16} \mathrm{~Hz}\right)$. The $v_0$ of metal is
238720
The wavelengths of two photons are $2000 \AA$ and $4000 \AA$ respectively. What is the ratio of their energies?
1 $1 / 4$
2 4
3 $1 / 2$
4 2
Explanation:
: Given that, $\lambda_1=2000 \AA, \lambda_2=4000 \AA$ We know that, $\therefore \quad \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ So, $\quad \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{hc}}{\lambda_1} \times \frac{\lambda_2}{\mathrm{hc}}$ $\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{4000 \AA}{2000 \AA} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=2 \end{aligned}$
VITEEE 2019
Structure of Atom
238721
The values of Planck's constant is $6.63 \times 10^{-34}$ Js. The velocity of light is $3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8 \times 10^{15}$
238722
The ratio of slopes of $K_{\max } \mathrm{vs} v$ and $v_0$ vs $v$ curves in the photoelectric effects gives $(v=$ frequency, $K_{\max }=$ maximum kinetic energy, $v_0$ = stopping potential)
1 The ratio of Planck's constant of electronic charge
2 Work function
3 Planck's constant
4 Charge of electron
Explanation:
: We know that, $\mathrm{h} v=\mathrm{h} v_0+\mathrm{e} v_0$ $\mathrm{e} v_0=\mathrm{h} v-\mathrm{h} v_0, \quad v_0=\frac{\mathrm{h}}{\mathrm{e}} v-\frac{\mathrm{h}}{\mathrm{e}} v_0$ On comparing this equation with the straight line equation i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $v_0$ vs $v$ is $\begin{aligned} & \text { (slope) })_1=\frac{\mathrm{h}}{\mathrm{e}} \\ & \mathrm{h} v=\mathrm{h} v_0+\mathrm{KE} \\ & \mathrm{KE}=\mathrm{h} v-\mathrm{h} v_0 \text { (constant) } \\ & \mathrm{y}=\mathrm{mx}+\left(-\mathrm{h} v_0\right) \end{aligned}$ Thus, slope of $\mathrm{K}_{\max }$ vs $v$ is $(\text { slope })_2=\mathrm{h}$ $\therefore \frac{(\text { slope })_2}{(\text { slope })_1}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$ (charge of electron)
VITEEE 2015
Structure of Atom
238723
A certain metal when irradiated by light $(r=$ $3.2 \times 10^{16} \mathrm{~Hz}$ ) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $(r=2.0 \times$ $\left.10^{16} \mathrm{~Hz}\right)$. The $v_0$ of metal is
